Antiderivatives are functions that reverse the process of differentiation. If \(F'(x)=f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\). These problems focus on using the power rule, rewriting radicals and fractions with exponents, and remembering to include the constant \(+C\).
Practice Problems
Find the antiderivative of each function.
\(\textbf{1)}\) \(\displaystyle f(x)= x+5 \)
The antiderivative is \(\displaystyle \frac{x^2}{2}+5x+C\)
\(\,\,\,\,\,\displaystyle \int (x+5)\,dx\)
\(\,\,\,\,\,\displaystyle \int x\,dx+\int 5\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^2}{2}+5x+C\)
\(\,\,\,\,\,\)The antiderivative is \(\displaystyle \frac{x^2}{2}+5x+C\)
\(\textbf{2)}\) \(\displaystyle f(x)= x^4-2x^2+5x \)
The antiderivative is \(\displaystyle \frac{x^5}{5}-\displaystyle \frac{2x^3}{3}+\displaystyle \frac{5x^2}{2}+C\)
\(\,\,\,\,\,\displaystyle \int \left(x^4-2x^2+5x\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int x^4\,dx-\int 2x^2\,dx+\int 5x\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^5}{5}-2\cdot\frac{x^3}{3}+5\cdot\frac{x^2}{2}+C\)
\(\,\,\,\,\,\displaystyle \frac{x^5}{5}-\frac{2x^3}{3}+\frac{5x^2}{2}+C\)
\(\,\,\,\,\,\)The antiderivative is \(\displaystyle \frac{x^5}{5}-\frac{2x^3}{3}+\frac{5x^2}{2}+C\)
\(\textbf{3)}\) \(\displaystyle f(x)= x^{1.5}+5x \)
The antiderivative is \(\displaystyle \frac{2x^{2.5}}{5}+\displaystyle \frac{5x^2}{2}+C\)
\(\,\,\,\,\,\displaystyle \int \left(x^{1.5}+5x\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int x^{1.5}\,dx+\int 5x\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{2.5}}{2.5}+\frac{5x^2}{2}+C\)
\(\,\,\,\,\,\displaystyle \frac{2x^{2.5}}{5}+\frac{5x^2}{2}+C\)
\(\,\,\,\,\,\)The antiderivative is \(\displaystyle \frac{2x^{2.5}}{5}+\frac{5x^2}{2}+C\)
\(\textbf{4)}\) \(\displaystyle f(x)= \displaystyle \frac{1}{x^5} \)
The antiderivative is \(\displaystyle \frac{-1}{4x^4}+C\)
\(\,\,\,\,\,\displaystyle \int \frac{1}{x^5}\,dx\)
\(\,\,\,\,\,\displaystyle \int x^{-5}\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{-4}}{-4}+C\)
\(\,\,\,\,\,\displaystyle -\frac{1}{4x^4}+C\)
\(\,\,\,\,\,\)The antiderivative is \(\displaystyle -\frac{1}{4x^4}+C\)
\(\textbf{5)}\) \(\displaystyle f(x)= \sqrt[3]{x}+2 \)
The antiderivative is \(\displaystyle \frac{3x^{4/3}}{4}+2x+C\)
\(\,\,\,\,\,\displaystyle \int \left(\sqrt[3]{x}+2\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(x^{1/3}+2\right)\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{4/3}}{4/3}+2x+C\)
\(\,\,\,\,\,\displaystyle \frac{3x^{4/3}}{4}+2x+C\)
\(\,\,\,\,\,\)The antiderivative is \(\displaystyle \frac{3x^{4/3}}{4}+2x+C\)
\(\textbf{6)}\) \(\displaystyle f(x)= \displaystyle \frac{x^3+5x^2-4}{\sqrt{x}} \)
The antiderivative is \(\displaystyle \frac{2x^{3.5}}{7}+2x^{2.5}-8\sqrt{x}+C\)
\(\,\,\,\,\,\displaystyle \int \frac{x^3+5x^2-4}{\sqrt{x}}\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(\frac{x^3}{x^{1/2}}+\frac{5x^2}{x^{1/2}}-\frac{4}{x^{1/2}}\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(x^{5/2}+5x^{3/2}-4x^{-1/2}\right)\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{7/2}}{7/2}+5\cdot\frac{x^{5/2}}{5/2}-4\cdot\frac{x^{1/2}}{1/2}+C\)
\(\,\,\,\,\,\displaystyle \frac{2x^{7/2}}{7}+2x^{5/2}-8\sqrt{x}+C\)
\(\,\,\,\,\,\)The antiderivative is \(\displaystyle \frac{2x^{3.5}}{7}+2x^{2.5}-8\sqrt{x}+C\)
\(\textbf{7)}\) \(\displaystyle f(x)= x^3{\sqrt{x}} \)
The antiderivative is \(\displaystyle \frac{2x^{4.5}}{9}+C\)
\(\,\,\,\,\,\displaystyle \int x^3\sqrt{x}\,dx\)
\(\,\,\,\,\,\displaystyle \int x^3x^{1/2}\,dx\)
\(\,\,\,\,\,\displaystyle \int x^{7/2}\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{9/2}}{9/2}+C\)
\(\,\,\,\,\,\displaystyle \frac{2x^{9/2}}{9}+C\)
\(\,\,\,\,\,\)The antiderivative is \(\displaystyle \frac{2x^{4.5}}{9}+C\)
\(\textbf{8)}\) \(\displaystyle f(x)= 0 \)
The antiderivative is \(C\)
\(\,\,\,\,\,\displaystyle \int 0\,dx\)
\(\,\,\,\,\,\text{Any constant has derivative }0.\)
\(\,\,\,\,\,\)The antiderivative is \(C\)
\(\textbf{9)}\) \(\displaystyle f(x)=7x^6\)
The antiderivative is \(x^7+C\)
\(\,\,\,\,\,\displaystyle \int 7x^6\,dx\)
\(\,\,\,\,\,\displaystyle 7\cdot\frac{x^7}{7}+C\)
\(\,\,\,\,\,\displaystyle x^7+C\)
\(\,\,\,\,\,\)The antiderivative is \(x^7+C\)
\(\textbf{10)}\) \(\displaystyle f(x)=4x^3-9x+2\)
The antiderivative is \(x^4-\frac{9x^2}{2}+2x+C\)
\(\,\,\,\,\,\displaystyle \int \left(4x^3-9x+2\right)\,dx\)
\(\,\,\,\,\,\displaystyle 4\cdot\frac{x^4}{4}-9\cdot\frac{x^2}{2}+2x+C\)
\(\,\,\,\,\,\displaystyle x^4-\frac{9x^2}{2}+2x+C\)
\(\,\,\,\,\,\)The antiderivative is \(x^4-\frac{9x^2}{2}+2x+C\)
\(\textbf{11)}\) \(\displaystyle f(x)=\sqrt{x}+x^2\)
The antiderivative is \(\frac{2x^{3/2}}{3}+\frac{x^3}{3}+C\)
\(\,\,\,\,\,\displaystyle \int \left(\sqrt{x}+x^2\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(x^{1/2}+x^2\right)\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{3/2}}{3/2}+\frac{x^3}{3}+C\)
\(\,\,\,\,\,\displaystyle \frac{2x^{3/2}}{3}+\frac{x^3}{3}+C\)
\(\,\,\,\,\,\)The antiderivative is \(\frac{2x^{3/2}}{3}+\frac{x^3}{3}+C\)
\(\textbf{12)}\) \(\displaystyle f(x)=\frac{6}{x^4}\)
The antiderivative is \(-\frac{2}{x^3}+C\)
\(\,\,\,\,\,\displaystyle \int \frac{6}{x^4}\,dx\)
\(\,\,\,\,\,\displaystyle \int 6x^{-4}\,dx\)
\(\,\,\,\,\,\displaystyle 6\cdot\frac{x^{-3}}{-3}+C\)
\(\,\,\,\,\,\displaystyle -2x^{-3}+C\)
\(\,\,\,\,\,\)The antiderivative is \(-\frac{2}{x^3}+C\)
\(\textbf{13)}\) \(\displaystyle f(x)=\frac{x^2+3x+1}{x}\)
The antiderivative is \(\frac{x^2}{2}+3x+\ln|x|+C\)
\(\,\,\,\,\,\displaystyle \int \frac{x^2+3x+1}{x}\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(x+3+\frac{1}{x}\right)\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^2}{2}+3x+\ln|x|+C\)
\(\,\,\,\,\,\)The antiderivative is \(\frac{x^2}{2}+3x+\ln|x|+C\)
\(\textbf{14)}\) \(\displaystyle f(x)=\frac{5x^2-4x+7}{x^2}\)
The antiderivative is \(5x-4\ln|x|-\frac{7}{x}+C\)
\(\,\,\,\,\,\displaystyle \int \frac{5x^2-4x+7}{x^2}\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(5-\frac{4}{x}+\frac{7}{x^2}\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(5-4x^{-1}+7x^{-2}\right)\,dx\)
\(\,\,\,\,\,\displaystyle 5x-4\ln|x|+7\cdot\frac{x^{-1}}{-1}+C\)
\(\,\,\,\,\,\displaystyle 5x-4\ln|x|-\frac{7}{x}+C\)
\(\,\,\,\,\,\)The antiderivative is \(5x-4\ln|x|-\frac{7}{x}+C\)
\(\textbf{15)}\) \(\displaystyle f(x)=3\sqrt[4]{x^3}\)
The antiderivative is \(\frac{12x^{7/4}}{7}+C\)
\(\,\,\,\,\,\displaystyle \int 3\sqrt[4]{x^3}\,dx\)
\(\,\,\,\,\,\displaystyle \int 3x^{3/4}\,dx\)
\(\,\,\,\,\,\displaystyle 3\cdot\frac{x^{7/4}}{7/4}+C\)
\(\,\,\,\,\,\displaystyle \frac{12x^{7/4}}{7}+C\)
\(\,\,\,\,\,\)The antiderivative is \(\frac{12x^{7/4}}{7}+C\)
\(\textbf{16)}\) \(\displaystyle f(x)=x-\frac{1}{x^3}\)
The antiderivative is \(\frac{x^2}{2}+\frac{1}{2x^2}+C\)
\(\,\,\,\,\,\displaystyle \int \left(x-\frac{1}{x^3}\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(x-x^{-3}\right)\,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^2}{2}-\frac{x^{-2}}{-2}+C\)
\(\,\,\,\,\,\displaystyle \frac{x^2}{2}+\frac{1}{2x^2}+C\)
\(\,\,\,\,\,\)The antiderivative is \(\frac{x^2}{2}+\frac{1}{2x^2}+C\)
\(\textbf{17)}\) \(\displaystyle f(x)=9x^8-4x^{-2}\)
The antiderivative is \(x^9+\frac{4}{x}+C\)
\(\,\,\,\,\,\displaystyle \int \left(9x^8-4x^{-2}\right)\,dx\)
\(\,\,\,\,\,\displaystyle 9\cdot\frac{x^9}{9}-4\cdot\frac{x^{-1}}{-1}+C\)
\(\,\,\,\,\,\displaystyle x^9+4x^{-1}+C\)
\(\,\,\,\,\,\)The antiderivative is \(x^9+\frac{4}{x}+C\)
\(\textbf{18)}\) \(\displaystyle f(x)=\frac{2}{\sqrt{x}}+x^3\)
The antiderivative is \(4\sqrt{x}+\frac{x^4}{4}+C\)
\(\,\,\,\,\,\displaystyle \int \left(\frac{2}{\sqrt{x}}+x^3\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(2x^{-1/2}+x^3\right)\,dx\)
\(\,\,\,\,\,\displaystyle 2\cdot\frac{x^{1/2}}{1/2}+\frac{x^4}{4}+C\)
\(\,\,\,\,\,\displaystyle 4\sqrt{x}+\frac{x^4}{4}+C\)
\(\,\,\,\,\,\)The antiderivative is \(4\sqrt{x}+\frac{x^4}{4}+C\)
\(\textbf{19)}\) \(\displaystyle f(x)=6x^5+\frac{3}{x}-\frac{2}{x^2}\)
The antiderivative is \(x^6+3\ln|x|+\frac{2}{x}+C\)
\(\,\,\,\,\,\displaystyle \int \left(6x^5+\frac{3}{x}-\frac{2}{x^2}\right)\,dx\)
\(\,\,\,\,\,\displaystyle \int \left(6x^5+3x^{-1}-2x^{-2}\right)\,dx\)
\(\,\,\,\,\,\displaystyle 6\cdot\frac{x^6}{6}+3\ln|x|-2\cdot\frac{x^{-1}}{-1}+C\)
\(\,\,\,\,\,\displaystyle x^6+3\ln|x|+\frac{2}{x}+C\)
\(\,\,\,\,\,\)The antiderivative is \(x^6+3\ln|x|+\frac{2}{x}+C\)
\(\textbf{20)}\) \(\displaystyle f(x)=12\)
The antiderivative is \(12x+C\)
\(\,\,\,\,\,\displaystyle \int 12\,dx\)
\(\,\,\,\,\,\displaystyle 12x+C\)
\(\,\,\,\,\,\)The antiderivative is \(12x+C\)
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