Integration by Substitution, also called u-substitution, is a method for rewriting a complicated integral in a simpler form. The goal is to choose a new variable \(u\) so that part of the integral becomes \(du\). This technique is especially useful with composite functions, powers, roots, trigonometric functions, exponential functions, and logarithmic expressions.
Notes
Practice Problems
Find the integral
\(\textbf{1)}\)\(\displaystyle \int (x^2+3)^3(2x) \,dx\)\(\text{The answer is } \frac{1}{4} (x^2+3)^4+C\)
\(\,\,\,\,\,u=x^2+3\)
\(\,\,\,\,\,du=2x \,dx\)
\(\,\,\,\,\,\displaystyle \int u^3 \,du\)
\(\,\,\,\,\,\displaystyle \frac{u^4}{4}+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{4} (x^2+3)^4+C\)
\(\,\,\,\,\,u=x^2+3\)
\(\,\,\,\,\,du=2x \,dx\)
\(\,\,\,\,\,\displaystyle \int u^3 \,du\)
\(\,\,\,\,\,\displaystyle \frac{u^4}{4}+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{4} (x^2+3)^4+C\)
\(\textbf{2)}\)\(\displaystyle \int x^2(5x^3+3)^2 \,dx\)\(\text{The answer is } \frac{1}{45}(5x^3+3)^3+C\)
\(\,\,\,\,\,u=5x^3+3\)
\(\,\,\,\,\,du=15x^2 \,dx\)
\(\,\,\,\,\,\frac{1}{15}du=x^2\, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{15}\int u^2 \,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{15}\cdot\frac{u^3}{3}+C\)
\(\,\,\,\,\,\displaystyle \frac{u^3}{45}+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{45}(5x^3+3)^3+C\)
\(\,\,\,\,\,u=5x^3+3\)
\(\,\,\,\,\,du=15x^2 \,dx\)
\(\,\,\,\,\,\frac{1}{15}du=x^2\, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{15}\int u^2 \,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{15}\cdot\frac{u^3}{3}+C\)
\(\,\,\,\,\,\displaystyle \frac{u^3}{45}+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{45}(5x^3+3)^3+C\)
\(\textbf{3)}\)\(\displaystyle \int \frac{\sin x}{\cos^4 x} \,dx\)\(\text{The answer is } \displaystyle \frac{1}{3 \cos^3 x }+C\)
\(\,\,\,\,\,u=\cos x\)
\(\,\,\,\,\,du=-\sin x \, dx\)
\(\,\,\,\,\,-du=\sin x \, dx\)
\(\,\,\,\,\,\displaystyle -\int \frac{1}{u^4} \,du\)
\(\,\,\,\,\,\displaystyle -\int u^{-4} \,du\)
\(\,\,\,\,\,\displaystyle -\frac{u^{-3}}{-3}+C\)
\(\,\,\,\,\,\displaystyle \frac{u^{-3}}{3}+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{1}{3 \cos^3 x }+C\)
\(\,\,\,\,\,u=\cos x\)
\(\,\,\,\,\,du=-\sin x \, dx\)
\(\,\,\,\,\,-du=\sin x \, dx\)
\(\,\,\,\,\,\displaystyle -\int \frac{1}{u^4} \,du\)
\(\,\,\,\,\,\displaystyle -\int u^{-4} \,du\)
\(\,\,\,\,\,\displaystyle -\frac{u^{-3}}{-3}+C\)
\(\,\,\,\,\,\displaystyle \frac{u^{-3}}{3}+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{1}{3 \cos^3 x }+C\)
\(\textbf{4)}\)\(\displaystyle \int x \sin (x^2) \,dx\)\(\text{The answer is } \displaystyle -\frac{1}{2} \cos(x^2)+C\)
\(\,\,\,\,\,u=x^2\)
\(\,\,\,\,\,du=2x \,dx\)
\(\,\,\,\,\,\frac{1}{2}du= x \, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} \int \sin u \,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} (-\cos u) +C\)
\(\,\,\,\,\,\)The answer is \(-\frac{1}{2} \cos(x^2)+C\)
\(\,\,\,\,\,u=x^2\)
\(\,\,\,\,\,du=2x \,dx\)
\(\,\,\,\,\,\frac{1}{2}du= x \, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} \int \sin u \,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} (-\cos u) +C\)
\(\,\,\,\,\,\)The answer is \(-\frac{1}{2} \cos(x^2)+C\)
\(\textbf{5)}\)\(\displaystyle \int \frac{8x^4}{\sqrt{x^5-2}} \,dx\)\(\text{The answer is } \displaystyle \frac{16}{5} \sqrt{x^5-2}+C\)
\(\,\,\,\,\,u=x^5-2\)
\(\,\,\,\,\,du=5x^4 \,dx\)
\(\,\,\,\,\,\frac{1}{5}du= x^4 \, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{5} \int \frac{8}{\sqrt{u}} \,du\)
\(\,\,\,\,\,\displaystyle \frac{8}{5} \int u^{-1/2} \,du\)
\(\,\,\,\,\,\displaystyle \frac{8}{5} \cdot 2u^{1/2} +C\)
\(\,\,\,\,\,\displaystyle \frac{16}{5} \sqrt{u} +C\)
\(\,\,\,\,\,\)The answer is \(\frac{16}{5} \sqrt{x^5-2}+C\)
\(\,\,\,\,\,u=x^5-2\)
\(\,\,\,\,\,du=5x^4 \,dx\)
\(\,\,\,\,\,\frac{1}{5}du= x^4 \, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{5} \int \frac{8}{\sqrt{u}} \,du\)
\(\,\,\,\,\,\displaystyle \frac{8}{5} \int u^{-1/2} \,du\)
\(\,\,\,\,\,\displaystyle \frac{8}{5} \cdot 2u^{1/2} +C\)
\(\,\,\,\,\,\displaystyle \frac{16}{5} \sqrt{u} +C\)
\(\,\,\,\,\,\)The answer is \(\frac{16}{5} \sqrt{x^5-2}+C\)
\(\textbf{6)}\)\(\displaystyle \int 5x^2 \sqrt{x^3+10} \,dx\)\(\text{The answer is } \displaystyle \frac{10}{9} (x^3+10)^{3/2}+C\)
\(\,\,\,\,\,u=x^3+10\)
\(\,\,\,\,\,du=3x^2 \,dx\)
\(\,\,\,\,\,\frac{1}{3}du= x^2 \, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{3} \int 5 \sqrt{u} \,du\)
\(\,\,\,\,\,\displaystyle \frac{5}{3} \int u^{1/2} \,du\)
\(\,\,\,\,\,\displaystyle \frac{5}{3} \cdot \frac{2}{3} u^{3/2} +C\)
\(\,\,\,\,\,\displaystyle \frac{10}{9} u^{3/2} +C\)
\(\,\,\,\,\,\)The answer is \(\frac{10}{9} (x^3+10)^{3/2}+C\)
\(\,\,\,\,\,u=x^3+10\)
\(\,\,\,\,\,du=3x^2 \,dx\)
\(\,\,\,\,\,\frac{1}{3}du= x^2 \, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{3} \int 5 \sqrt{u} \,du\)
\(\,\,\,\,\,\displaystyle \frac{5}{3} \int u^{1/2} \,du\)
\(\,\,\,\,\,\displaystyle \frac{5}{3} \cdot \frac{2}{3} u^{3/2} +C\)
\(\,\,\,\,\,\displaystyle \frac{10}{9} u^{3/2} +C\)
\(\,\,\,\,\,\)The answer is \(\frac{10}{9} (x^3+10)^{3/2}+C\)
\(\textbf{7)}\)\(\displaystyle \int \frac{\sec^2 (\frac{1}{x^6})}{x^7} \,dx\)\(\text{The answer is } \displaystyle -\frac{1}{6} \tan \left(\frac{1}{x^6}\right)+C\)
\(\,\,\,\,\,u=\displaystyle \frac{1}{x^6}\)
\(\,\,\,\,\,u=\displaystyle x^{-6}\)
\(\,\,\,\,\,du=-6x^{-7} \,dx\)
\(\,\,\,\,\,\displaystyle -\frac{1}{6}du=x^{-7} \, dx\)
\(\,\,\,\,\,\displaystyle -\frac{1}{6} \int \sec^2 u \,du\)
\(\,\,\,\,\,\displaystyle -\frac{1}{6} \tan u +C\)
\(\,\,\,\,\,\)The answer is \(-\frac{1}{6} \tan \left(\frac{1}{x^6}\right)+C\)
\(\,\,\,\,\,u=\displaystyle \frac{1}{x^6}\)
\(\,\,\,\,\,u=\displaystyle x^{-6}\)
\(\,\,\,\,\,du=-6x^{-7} \,dx\)
\(\,\,\,\,\,\displaystyle -\frac{1}{6}du=x^{-7} \, dx\)
\(\,\,\,\,\,\displaystyle -\frac{1}{6} \int \sec^2 u \,du\)
\(\,\,\,\,\,\displaystyle -\frac{1}{6} \tan u +C\)
\(\,\,\,\,\,\)The answer is \(-\frac{1}{6} \tan \left(\frac{1}{x^6}\right)+C\)
\(\textbf{8)}\)\(\displaystyle \int \sec^2 x \tan^2 x \,dx\)\(\text{The answer is } \displaystyle \frac{1}{3} \tan^3(x) +C\)
\(\,\,\,\,\,u=\tan x\)
\(\,\,\,\,\,du= \sec^2 x \,dx\)
\(\,\,\,\,\,\displaystyle \int u^2 \,du\)
\(\,\,\,\,\,\displaystyle \frac{u^3}{3} +C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{3} \tan^3(x) +C\)
\(\,\,\,\,\,u=\tan x\)
\(\,\,\,\,\,du= \sec^2 x \,dx\)
\(\,\,\,\,\,\displaystyle \int u^2 \,du\)
\(\,\,\,\,\,\displaystyle \frac{u^3}{3} +C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{3} \tan^3(x) +C\)
\(\textbf{9)}\)\(\displaystyle \int e^x \sqrt{15+e^x} \,dx\)\(\text{The answer is } \displaystyle \frac{2}{3} (15+e^x )^{3/2} +C\)
\(\,\,\,\,\,u=15+e^x\)
\(\,\,\,\,\,du= e^x \,dx\)
\(\,\,\,\,\,\displaystyle \int \sqrt{u} \,du\)
\(\,\,\,\,\,\displaystyle \int u^{1/2} \,du\)
\(\,\,\,\,\,\displaystyle \frac{2u^{3/2}}{3} +C\)
\(\,\,\,\,\,\)The answer is \(\frac{2}{3} (15+e^x )^{3/2} +C\)
\(\,\,\,\,\,u=15+e^x\)
\(\,\,\,\,\,du= e^x \,dx\)
\(\,\,\,\,\,\displaystyle \int \sqrt{u} \,du\)
\(\,\,\,\,\,\displaystyle \int u^{1/2} \,du\)
\(\,\,\,\,\,\displaystyle \frac{2u^{3/2}}{3} +C\)
\(\,\,\,\,\,\)The answer is \(\frac{2}{3} (15+e^x )^{3/2} +C\)
\(\textbf{10)}\)\(\displaystyle \int e^{\cos 5 \theta} \sin 5\theta \,d\theta\)\(\text{The answer is } \displaystyle -\frac{1}{5} e^{\cos(5\theta)} +C\)
\(\,\,\,\,\,u=\cos (5 \theta)\)
\(\,\,\,\,\,du= -5\sin (5\theta) \,d\theta\)
\(\,\,\,\,\,-\frac{1}{5}du= \sin (5\theta) \, d\theta\)
\(\,\,\,\,\,\displaystyle -\frac{1}{5} \int e^u \,du\)
\(\,\,\,\,\,\displaystyle -\frac{1}{5}e^u +C\)
\(\,\,\,\,\,\)The answer is \(-\frac{1}{5} e^{\cos(5\theta)} +C\)
\(\,\,\,\,\,u=\cos (5 \theta)\)
\(\,\,\,\,\,du= -5\sin (5\theta) \,d\theta\)
\(\,\,\,\,\,-\frac{1}{5}du= \sin (5\theta) \, d\theta\)
\(\,\,\,\,\,\displaystyle -\frac{1}{5} \int e^u \,du\)
\(\,\,\,\,\,\displaystyle -\frac{1}{5}e^u +C\)
\(\,\,\,\,\,\)The answer is \(-\frac{1}{5} e^{\cos(5\theta)} +C\)
\(\textbf{11)}\)\(\displaystyle \int \frac{\sin (\ln(4x))}{x} \,dx\)\(\text{The answer is } \displaystyle -\cos(\ln(4x))+C\)
\(\,\,\,\,\,u=\ln(4x)\)
\(\,\,\,\,\,du=\displaystyle \frac{1}{x} \,dx\)
\(\,\,\,\,\,\displaystyle \int \sin u \,du\)
\(\,\,\,\,\,\displaystyle -\cos u +C\)
\(\,\,\,\,\,\)The answer is \(-\cos(\ln(4x))+C\)
\(\,\,\,\,\,u=\ln(4x)\)
\(\,\,\,\,\,du=\displaystyle \frac{1}{x} \,dx\)
\(\,\,\,\,\,\displaystyle \int \sin u \,du\)
\(\,\,\,\,\,\displaystyle -\cos u +C\)
\(\,\,\,\,\,\)The answer is \(-\cos(\ln(4x))+C\)
\(\textbf{12)}\)\(\displaystyle \int x e^{x^2} \,dx\)\(\text{The answer is } \displaystyle \frac{1}{2} e^{x^2} +C\)
\(\,\,\,\,\,u=x^2\)
\(\,\,\,\,\,du=2x \,dx\)
\(\,\,\,\,\,\frac{1}{2}du= x \, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} \int e^u \,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} e^u +C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{2} e^{x^2} +C\)
\(\,\,\,\,\,u=x^2\)
\(\,\,\,\,\,du=2x \,dx\)
\(\,\,\,\,\,\frac{1}{2}du= x \, dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} \int e^u \,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} e^u +C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{2} e^{x^2} +C\)
\(\textbf{13)}\)\(\displaystyle \int \frac{x}{\sqrt{x^2-7}} \,dx\)\(\text{The answer is } \sqrt{x^2-7}+C\)
\(\,\,\,\,\,u=x^2-7\)
\(\,\,\,\,\,du=2x \,dx\)
\(\,\,\,\,\,\frac{1}{2}du=x \,dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} \int \frac{1}{\sqrt{u}} \,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\int u^{-1/2}\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\cdot 2u^{1/2}+C\)
\(\,\,\,\,\,\displaystyle \sqrt{u}+C\)
\(\,\,\,\,\,\)The answer is \(\sqrt{x^2-7}+C\)
\(\,\,\,\,\,u=x^2-7\)
\(\,\,\,\,\,du=2x \,dx\)
\(\,\,\,\,\,\frac{1}{2}du=x \,dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2} \int \frac{1}{\sqrt{u}} \,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\int u^{-1/2}\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\cdot 2u^{1/2}+C\)
\(\,\,\,\,\,\displaystyle \sqrt{u}+C\)
\(\,\,\,\,\,\)The answer is \(\sqrt{x^2-7}+C\)
\(\textbf{14)}\)\(\displaystyle \int \frac{3x^2}{x^3+5}\,dx\)\(\text{The answer is } \ln|x^3+5|+C\)
\(\,\,\,\,\,u=x^3+5\)
\(\,\,\,\,\,du=3x^2\,dx\)
\(\,\,\,\,\,\displaystyle \int \frac{1}{u}\,du\)
\(\,\,\,\,\,\displaystyle \ln|u|+C\)
\(\,\,\,\,\,\)The answer is \(\ln|x^3+5|+C\)
\(\,\,\,\,\,u=x^3+5\)
\(\,\,\,\,\,du=3x^2\,dx\)
\(\,\,\,\,\,\displaystyle \int \frac{1}{u}\,du\)
\(\,\,\,\,\,\displaystyle \ln|u|+C\)
\(\,\,\,\,\,\)The answer is \(\ln|x^3+5|+C\)
\(\textbf{15)}\)\(\displaystyle \int \cos(7x)\,dx\)\(\text{The answer is } \frac{1}{7}\sin(7x)+C\)
\(\,\,\,\,\,u=7x\)
\(\,\,\,\,\,du=7\,dx\)
\(\,\,\,\,\,\frac{1}{7}du=dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{7}\int \cos u\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{7}\sin u+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{7}\sin(7x)+C\)
\(\,\,\,\,\,u=7x\)
\(\,\,\,\,\,du=7\,dx\)
\(\,\,\,\,\,\frac{1}{7}du=dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{7}\int \cos u\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{7}\sin u+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{7}\sin(7x)+C\)
\(\textbf{16)}\)\(\displaystyle \int \frac{1}{2x+1}\,dx\)\(\text{The answer is } \frac{1}{2}\ln|2x+1|+C\)
\(\,\,\,\,\,u=2x+1\)
\(\,\,\,\,\,du=2\,dx\)
\(\,\,\,\,\,\frac{1}{2}du=dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\int \frac{1}{u}\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\ln|u|+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{2}\ln|2x+1|+C\)
\(\,\,\,\,\,u=2x+1\)
\(\,\,\,\,\,du=2\,dx\)
\(\,\,\,\,\,\frac{1}{2}du=dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\int \frac{1}{u}\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\ln|u|+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{2}\ln|2x+1|+C\)
\(\textbf{17)}\)\(\displaystyle \int \frac{x}{x^2+4}\,dx\)\(\text{The answer is } \frac{1}{2}\ln(x^2+4)+C\)
\(\,\,\,\,\,u=x^2+4\)
\(\,\,\,\,\,du=2x\,dx\)
\(\,\,\,\,\,\frac{1}{2}du=x\,dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\int \frac{1}{u}\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\ln|u|+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{2}\ln(x^2+4)+C\)
\(\,\,\,\,\,u=x^2+4\)
\(\,\,\,\,\,du=2x\,dx\)
\(\,\,\,\,\,\frac{1}{2}du=x\,dx\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\int \frac{1}{u}\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\ln|u|+C\)
\(\,\,\,\,\,\)The answer is \(\frac{1}{2}\ln(x^2+4)+C\)
\(\textbf{18)}\)\(\displaystyle \int_0^2 x(x^2+1)^4\,dx\)\(\text{The answer is } \frac{3124}{10}=\frac{1562}{5}\)
\(\,\,\,\,\,u=x^2+1\)
\(\,\,\,\,\,du=2x\,dx\)
\(\,\,\,\,\,\frac{1}{2}du=x\,dx\)
\(\,\,\,\,\,x=0 \rightarrow u=1\)
\(\,\,\,\,\,x=2 \rightarrow u=5\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\int_1^5 u^4\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\left[\frac{u^5}{5}\right]_1^5\)
\(\,\,\,\,\,\displaystyle \frac{1}{10}(5^5-1^5)\)
\(\,\,\,\,\,\)The answer is \(\frac{3124}{10}=\frac{1562}{5}\)
\(\,\,\,\,\,u=x^2+1\)
\(\,\,\,\,\,du=2x\,dx\)
\(\,\,\,\,\,\frac{1}{2}du=x\,dx\)
\(\,\,\,\,\,x=0 \rightarrow u=1\)
\(\,\,\,\,\,x=2 \rightarrow u=5\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\int_1^5 u^4\,du\)
\(\,\,\,\,\,\displaystyle \frac{1}{2}\left[\frac{u^5}{5}\right]_1^5\)
\(\,\,\,\,\,\displaystyle \frac{1}{10}(5^5-1^5)\)
\(\,\,\,\,\,\)The answer is \(\frac{3124}{10}=\frac{1562}{5}\)
\(\textbf{19)}\)\(\displaystyle \int_1^3 \frac{2x}{x^2+1}\,dx\)\(\text{The answer is } \ln 5\)
\(\,\,\,\,\,u=x^2+1\)
\(\,\,\,\,\,du=2x\,dx\)
\(\,\,\,\,\,x=1 \rightarrow u=2\)
\(\,\,\,\,\,x=3 \rightarrow u=10\)
\(\,\,\,\,\,\displaystyle \int_2^{10}\frac{1}{u}\,du\)
\(\,\,\,\,\,\displaystyle \left[\ln|u|\right]_2^{10}\)
\(\,\,\,\,\,\displaystyle \ln 10-\ln 2\)
\(\,\,\,\,\,\)The answer is \(\ln 5\)
\(\,\,\,\,\,u=x^2+1\)
\(\,\,\,\,\,du=2x\,dx\)
\(\,\,\,\,\,x=1 \rightarrow u=2\)
\(\,\,\,\,\,x=3 \rightarrow u=10\)
\(\,\,\,\,\,\displaystyle \int_2^{10}\frac{1}{u}\,du\)
\(\,\,\,\,\,\displaystyle \left[\ln|u|\right]_2^{10}\)
\(\,\,\,\,\,\displaystyle \ln 10-\ln 2\)
\(\,\,\,\,\,\)The answer is \(\ln 5\)
\(\textbf{20)}\)\(\displaystyle \int \frac{e^{1/x}}{x^2}\,dx\)\(\text{The answer is } -e^{1/x}+C\)
\(\,\,\,\,\,u=\frac{1}{x}\)
\(\,\,\,\,\,u=x^{-1}\)
\(\,\,\,\,\,du=-x^{-2}\,dx\)
\(\,\,\,\,\,-du=\frac{1}{x^2}\,dx\)
\(\,\,\,\,\,\displaystyle -\int e^u\,du\)
\(\,\,\,\,\,\displaystyle -e^u+C\)
\(\,\,\,\,\,\)The answer is \(-e^{1/x}+C\)
\(\,\,\,\,\,u=\frac{1}{x}\)
\(\,\,\,\,\,u=x^{-1}\)
\(\,\,\,\,\,du=-x^{-2}\,dx\)
\(\,\,\,\,\,-du=\frac{1}{x^2}\,dx\)
\(\,\,\,\,\,\displaystyle -\int e^u\,du\)
\(\,\,\,\,\,\displaystyle -e^u+C\)
\(\,\,\,\,\,\)The answer is \(-e^{1/x}+C\)
See Related Pages\(\)
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\(\,\,\,\,\,\,\,\,\displaystyle\frac{b-a}{2n}\left[f(a)+2f(x_1)+2f(x_2)+…+2fx_{n-1}+f(b)\right]…\)
\(\bullet\text{ Properties of Integrals}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{a}^{b}cf(x) \, dx=c\displaystyle \int_{a}^{b}f(x) \,dx…\)
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\(\bullet\text{ Indefinite Integrals- Trig Functions}\)
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\(\bullet\text{ Definite Integrals}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{5}^{7} x^3 \, dx…\)
\(\bullet\text{ Integration by Substitution}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int (x^2+3)^3(2x) \,dx…\)
\(\bullet\text{ Area of Region Between Two Curves}\)
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\(\bullet\text{ Arc Length}\)
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\(\bullet\text{ Volume by Cross Sections}\)
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