Indefinite integrals, also called antiderivatives, are used to reverse the process of differentiation. The power rule for integration says to add 1 to the exponent and then divide by the new exponent. These problems focus on rewriting expressions with exponents first, then applying the power rule and adding \(+C\).
Notes
Practice Problems
Find each indefinite integral
\(\textbf{1)}\) \(\displaystyle \int x+5 \,dx\)
The answer 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 answer is \(\displaystyle \frac{x^2}{2}+5x+C\)
\(\textbf{2)}\) \(\displaystyle \int x^4-2x^2+5x \,dx\)
The answer is \(\displaystyle \frac{x^5}{5}-\displaystyle \frac{2x^3}{3}+\displaystyle \frac{5x^2}{2}+C\)
\(\,\,\,\,\,\displaystyle \int x^4-2x^2+5x \,dx\)
\(\,\,\,\,\,\displaystyle \int x^4 \,dx-\int 2x^2 \,dx+\int 5x \,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^5}{5}-2\cdot\frac{x^3}{3}+5\cdot\frac{x^2}{2}+C\)
\(\,\,\,\,\,\displaystyle \frac{x^5}{5}-\frac{2x^3}{3}+\frac{5x^2}{2}+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{x^5}{5}-\frac{2x^3}{3}+\frac{5x^2}{2}+C\)
\(\textbf{3)}\) \(\displaystyle \int x^{1.5}+5x \,dx\)
The answer is \(\displaystyle \frac{2x^{2.5}}{5}+\displaystyle \frac{5x^2}{2}+C\)
\(\,\,\,\,\,\displaystyle \int x^{1.5}+5x \,dx\)
\(\,\,\,\,\,\displaystyle \int x^{1.5} \,dx+\int 5x \,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{2.5}}{2.5}+\frac{5x^2}{2}+C\)
\(\,\,\,\,\,\displaystyle \frac{2x^{2.5}}{5}+\frac{5x^2}{2}+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{2x^{2.5}}{5}+\frac{5x^2}{2}+C\)
\(\textbf{4)}\) \(\displaystyle \int \displaystyle \frac{1}{x^5} \,dx\)
The answer 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 answer is \(\displaystyle -\frac{1}{4x^4}+C\)
\(\textbf{5)}\) \(\displaystyle \int \sqrt[3]{x}+2 \,dx\)
The answer is \(\displaystyle \frac{3x^{4/3}}{4}+2x+C\)
\(\,\,\,\,\,\displaystyle \int \sqrt[3]{x}+2 \,dx\)
\(\,\,\,\,\,\displaystyle \int x^{1/3}+2 \,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{4/3}}{4/3}+2x+C\)
\(\,\,\,\,\,\displaystyle \frac{3x^{4/3}}{4}+2x+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{3x^{4/3}}{4}+2x+C\)
\(\textbf{6)}\) \(\displaystyle \int \displaystyle \frac{x^3+5x^2-4}{\sqrt{x}} \,dx\)
The answer 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 answer is \(\displaystyle \frac{2x^{3.5}}{7}+2x^{2.5}-8\sqrt{x}+C\)
\(\textbf{7)}\)\(\displaystyle \int x^3{\sqrt{x}} \,dx\)
The answer 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 answer is \(\displaystyle \frac{2x^{4.5}}{9}+C\)
\(\textbf{8)}\)\(\displaystyle \int dx\)
The answer is \( x + C \)
\(\,\,\,\,\,\displaystyle \int dx\)
\(\,\,\,\,\,\displaystyle \int 1 \,dx\)
\(\,\,\,\,\,\displaystyle x+C\)
\(\,\,\,\,\,\)The answer is \(x+C\)
\(\textbf{9)}\) \(\displaystyle \int \frac{1}{x} \,dx\)
The answer is \(\displaystyle \ln|x| + C\)
\(\,\,\,\,\,\text{Since } \frac{d}{dx} \ln|x| = \frac{1}{x}, \text{ we have } \int \frac{1}{x} \,dx = \ln|x| + C.\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \ln|x|+C\)
\(\textbf{10)}\) \(\displaystyle \int 7x^6 \,dx\)
The answer is \(x^7+C\)
\(\,\,\,\,\,\displaystyle \int 7x^6 \,dx\)
\(\,\,\,\,\,\displaystyle 7\cdot\frac{x^7}{7}+C\)
\(\,\,\,\,\,\)The answer is \(x^7+C\)
\(\textbf{11)}\) \(\displaystyle \int 4x^3-9x+2 \,dx\)
The answer is \(x^4-\frac{9x^2}{2}+2x+C\)
\(\,\,\,\,\,\displaystyle \int 4x^3-9x+2 \,dx\)
\(\,\,\,\,\,\displaystyle 4\cdot\frac{x^4}{4}-9\cdot\frac{x^2}{2}+2x+C\)
\(\,\,\,\,\,\displaystyle x^4-\frac{9x^2}{2}+2x+C\)
\(\,\,\,\,\,\)The answer is \(x^4-\frac{9x^2}{2}+2x+C\)
\(\textbf{12)}\) \(\displaystyle \int \sqrt{x}+x^2 \,dx\)
The answer is \(\frac{2x^{3/2}}{3}+\frac{x^3}{3}+C\)
\(\,\,\,\,\,\displaystyle \int \sqrt{x}+x^2 \,dx\)
\(\,\,\,\,\,\displaystyle \int x^{1/2}+x^2 \,dx\)
\(\,\,\,\,\,\displaystyle \frac{x^{3/2}}{3/2}+\frac{x^3}{3}+C\)
\(\,\,\,\,\,\displaystyle \frac{2x^{3/2}}{3}+\frac{x^3}{3}+C\)
\(\,\,\,\,\,\)The answer is \(\frac{2x^{3/2}}{3}+\frac{x^3}{3}+C\)
\(\textbf{13)}\) \(\displaystyle \int \frac{6}{x^4} \,dx\)
The answer 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 answer is \(-\frac{2}{x^3}+C\)
\(\textbf{14)}\) \(\displaystyle \int \frac{x^2+3x+1}{x} \,dx\)
The answer 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 answer is \(\frac{x^2}{2}+3x+\ln|x|+C\)
\(\textbf{15)}\) \(\displaystyle \int \frac{5x^2-4x+7}{x^2} \,dx\)
The answer 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 answer is \(5x-4\ln|x|-\frac{7}{x}+C\)
\(\textbf{16)}\) \(\displaystyle \int 3\sqrt[4]{x^3} \,dx\)
The answer 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 answer is \(\frac{12x^{7/4}}{7}+C\)
\(\textbf{17)}\) \(\displaystyle \int \left(x-\frac{1}{x^3}\right) \,dx\)
The answer 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 answer is \(\frac{x^2}{2}+\frac{1}{2x^2}+C\)
\(\textbf{18)}\) \(\displaystyle \int \left(9x^8-4x^{-2}\right) \,dx\)
The answer 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 answer is \(x^9+\frac{4}{x}+C\)
\(\textbf{19)}\) \(\displaystyle \int \left(\frac{2}{\sqrt{x}}+x^3\right) \,dx\)
The answer 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 answer is \(4\sqrt{x}+\frac{x^4}{4}+C\)
\(\textbf{20)}\) \(\displaystyle \int \left(6x^5+\frac{3}{x}-\frac{2}{x^2}\right) \,dx\)
The answer is \(x^6+3\ln|x|+\frac{2}{x}+C\)
\(\,\,\,\,\,\displaystyle \int \left(6x^5+\frac{3}{x}-\frac{2}{x^2}\right) \,dx\)
\(\,\,\,\,\,\displaystyle \int \left(6x^5+3x^{-1}-2x^{-2}\right) \,dx\)
\(\,\,\,\,\,\displaystyle 6\cdot\frac{x^6}{6}+3\ln|x|-2\cdot\frac{x^{-1}}{-1}+C\)
\(\,\,\,\,\,\displaystyle x^6+3\ln|x|+2x^{-1}+C\)
\(\,\,\,\,\,\)The answer is \(x^6+3\ln|x|+\frac{2}{x}+C\)
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