Methods of Integration (College Board AP® Calculus BC): Exam Questions

$\frac{1}{2}{\int }_2^3\cos udu$

$2{\int }_2^3\cos udu$

${\int }_4^9\cos udu$

$2{\int }_4^9\cos udu$

$x-\ln \left|x\right|+C$

$-x-\ln \left|x-1\right|+C$

$x+\ln \left|x-1\right|+C$

$\frac{x^2}{x^2-2x}+c$

$-2\left(\sqrt{2}+1\right)$

$-2\left(\sqrt{2}-1\right)$

$2\left(\sqrt{2}-1\right)$

$2\left(\sqrt{2}+1\right)$

$\arcsin \left(x-3\right)+\mathrm{C}$

$\mathrm{arc}\tan \left(x-3\right)+\mathrm{C}$

$\ln \left|x^2-6x+10\right|+C$

$\frac{1}{{\left(x^2-6x+10\right)}^2}+C$

$\frac{3x^3+8x^2+12x+24}{6x+12}+C$

$\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+2x+C$

$\begin{array}{rcl}& & \frac{1}{3}{x}^3+\frac{1}{2}{x}^2+x+\ln \left|x+1\right|+C\end{array}$

$\left(\frac{1}{4}{x}^4+\frac{2}{3}{x}^3+x^2+2x\right)·\ln \left|x+1\right|+C$

$\frac{-1}{8{\left(x^4-9\right)}^2}+C$

$\frac{1}{4}\ln \left|x^4-9\right|+C$

$4\ln \left|x^4-9\right|+C$

$\frac{1}{3}\arctan \left(\frac{x^2}{3}\right)+C$

$-\frac{1}{4}\cos \left(x^4\right)+C$

$\frac{1}{4}\cos \left(x^4\right)+C$

$-\frac{x^3}{4}\cos \left(x^4\right)+C$

$-\frac{x^4}{4}\cos \left(\frac{x^5}{5}\right)+C$

$\begin{array}{rcl}& & \frac{1}{30}{\left(5x^3+1\right)}^{\frac{1}{2}}+C\end{array}$

$\begin{array}{rcl}& & \frac{2}{15}{\left(5x^3+1\right)}^{\frac{1}{2}}+C\end{array}$

$\begin{array}{rcl}& & \frac{2}{3}{x}^3{\left(5x^3+1\right)}^{\frac{1}{2}}+C\end{array}$

$\begin{array}{rcl}& & \frac{2}{45}{\left(5x^3+1\right)}^{\frac{3}{2}}+C\end{array}$

$\frac{1}{4}{x}^4\left(\ln \left(3x\right)-3\right)+C$

$\frac{1}{16}{x}^4\left(4\ln \left(3x\right)-3\right)+C$

$\frac{1}{16}{x}^4\left(4\ln \left(3x\right)-1\right)+C$

$\frac{1}{3}{x}^3\left(3\ln \left(3x\right)-1\right)+C$

$\ln |x-2|-2\ln |x-3|+C$

$2\ln |x-3|+\ln |x-2|+C$

$2\ln |x-3|-\ln |x-2|+C$

$\frac{1}{2}\ln |x^2-5x+6|+C$

$-2$

$1$

$2$

$\infty$

$\frac{1}{5}\arcsin \left(5x\right)+C$

$\frac{x}{5}\arcsin \left(5x\right)+C$

$-\frac{1}{25}\sqrt{1-25x^2}+C$

$-\frac{1}{25}\ln \sqrt{1-25x^2}+C$

${\int }_1^{4}f\left(t\right)dt=3$

${\int }_1^{4}f\left(t\right)dt=27$

${\int }_9^{36}f\left(t\right)dt=3$

${\int }_9^{36}f\left(t\right)dt=27$

$\begin{array}{rcl}& & \frac{F\left(3\right)}{6}-\frac{F\left(2\right)}{6}\end{array}$

$\begin{array}{rcl}& & 3F\left(3\right)-3F\left(2\right)\end{array}$

$\begin{array}{rcl}& & \frac{F\left(81\right)}{6}-\frac{F\left(16\right)}{6}\end{array}$

$\begin{array}{rcl}& & 3F\left(81\right)-3F\left(16\right)\end{array}$

$\cos x\sqrt{1-\cos x}+C$

$\left(\cos x-1\right)\sqrt{1-\cos x}+C$

$\frac{3}{2}\left(1-\cos x\right)+C$

$\begin{array}{rcl}& & \frac{2}{3}{\left(1-\cos x\right)}^{\frac{3}{2}}+C\end{array}$

$\mathrm{arc}\sin \left(\frac{x-1}{2}\right)+C$

$2\mathrm{arc}\sin \left(\frac{x-1}{2}\right)+C$

$\mathrm{arc}\sin \left(x-1\right)+C$

$2\mathrm{arc}\sin \left(x-1\right)+C$

$\begin{array}{rcl}& & \frac{1}{4}{x}^4-\frac{2}{3}{x}^3+\frac{1}{2}{x}^2-7\ln \left|x-2\right|+C\end{array}$

$\begin{array}{rcl}& & \frac{1}{4}{x}^4+\frac{1}{2}{x}^2+2x-3\ln \left|x-2\right|+C\end{array}$

$\begin{array}{rcl}& & x^4+2x^2+8x-12\ln \left|x-2\right|+C\end{array}$

$\begin{array}{rcl}& & 3x^4-8x^3+6x^2-84\ln \left|x-2\right|+C\end{array}$

$\begin{array}{rcl}& & \frac{1}{16}{e}^{4x}\left(4x-1\right)+C\end{array}$

$\frac{1}{32}{e}^{4x}\left(8x^2+4x+1\right)+C$

$\frac{1}{32}{e}^{4x}\left(8x^2-4x+1\right)+C$

$\frac{1}{32}{e}^{4x}\left(8x^2-4x-1\right)+C$

$x^2+6\ln \left|x\right|-8\ln |x-1|+C$

$x^2-6\ln \left|x\right|+8\ln |x-1|+C$

$x^2+8\ln \left|x\right|-6\ln |x-1|+C$

$6\ln \left|x\right|-8\ln |x-1|+C$

0

2

4

The integral is divergent