Calculus/Integration techniques/Integration by Parts/Solutions

From Wikibooks, open books for an open world
Jump to: navigation, search
1. \int -4\ln(x)dx

Let

\begin{array}{ll}
u=\ln(x) & \qquad du=\frac{dx}{x}\\
v=-4x & \qquad dv=-4dx
\end{array}

Then

\begin{align}\int-4\ln(x)dx&=-4x\ln(x)-\int-4dx\\
&=\mathbf{-4x\ln(x)+4x+C}\end{align}
2. \int(-7x+38)\cos(x)dx

Let

\begin{array}{ll}
u=-7x+38 & \qquad du=-7dx\\
v=\sin(x) & \qquad dv=\cos(x)dx
\end{array}

Then

\begin{align}\int(-7x+38)\cos(x)dx&=(-7x+38)\sin(x)+7\int\sin(x)dx\\
&=\mathbf{(-7x+38)\sin(x)-7\cos(x)+C}\end{align}
3. \int_0^\frac{\pi}{2}(-6x+45)\cos(x)dx

Let

\begin{array}{ll}
u=-6x+45 & \qquad du=-6dx\\
v=\sin(x) & \qquad dv=\cos(x)dx
\end{array}

Then

\begin{align}\int_{0}^{\frac{\pi}{2}}(-6x+45)\cos(x)dx&=((-6x+45)\sin(x))\bigr|_{0}^{\frac{\pi}{2}}+6\int_{0}^{\frac{\pi}{2}}\sin(x)dx\\
&=(-3\pi+45)-6\cos(x)\bigr|_{0}^{\frac{\pi}{2}}\\
&=(-3\pi+45)+6\\
&=\mathbf{-3\pi+51}\end{align}
4. \int(5x+1)(x-6)^4 dx

Let

\begin{array}{ll}
u=5x+1 & \qquad du=5dx\\
v=\frac{(x-6)^{5}}{5} & \qquad dv=(x-6)^{4}dx
\end{array}

Then

\begin{align}\int(5x+1)(x-6)^{4}dx&=\frac{(5x+1)(x-6)^{5}}{5}-\int(x-6)^{5}dx\\
&=\mathbf{\frac{(5x+1)(x-6)^{5}}{5}-\frac{(x-6)^{6}}{6}+C}\end{align}
5. \int_{-1}^1 (2x+8)^3(-x+2)dx

Let

\begin{array}{ll}
u=-x+2 & \qquad du=-dx\\
v=\frac{(2x+8)^{4}}{8} & \qquad dv=(2x+8)^{3}dx
\end{array}

Then

\begin{align}\int_{-1}^{1}(2x+8)^{3}(-x+2)dx&=\frac{(-x+2)(2x+8)^{4}}{8}\Biggr|_{-1}^{1}+\int_{-1}^{1}\frac{(2x+8)^{4}}{8}dx\\
&=\frac{10^{4}-(3\cdot6^{4})}{8}+\frac{1}{8}\frac{(2x+8)^{5}}{10}\Biggr|_{-1}^{1}\\
&=\frac{10^{4}-3\cdot6^{4}}{8}+\frac{1}{80}(10^{5}-6^{5})\\
&=\frac{10^{5}-30\cdot6^{4}+10^{5}-6^{5}}{80}\\
&=\frac{2\cdot10^{5}-6^{4}(30+6)}{80}\\
&=\frac{2\cdot10^{5}-6^{6}}{80}\\
&=\mathbf{\frac{9584}{5}}\end{align}