Integration by parts

Integration by parts is the product rule $(fg)'=f'g+fg'$ integrated and rearranged.

The key decision is which factor to differentiate: choose the one that gets simpler ($x$ or $\log x$) as $f$. For $x\log x$, differentiate $\log x$.

Examples

1. 01What is the $uv$ part?$\displaystyle \begin{gathered} \displaystyle \int x e^x \, dx \\[6pt] \displaystyle (u=x,\; v'=e^{x} \;\Rightarrow\; v=e^{x}) \end{gathered}$AnswerHide

   Answer$\displaystyle x e^x$

   Steps

   1. $v$ is the integral of $v'$; multiply it by $u$
   2. $\displaystyle x \cdot e^{x}$$\displaystyle = x e^x$
2. 02What is $\int u'v\,dx$? (omit $C$)$\displaystyle \int x \log_e x \, dx = \frac{x^{2}}{2} \log_e x - \int u'v \, dx$AnswerHide

   Answer$\displaystyle \frac{x^{2}}{4}$

   Steps

   1. $u=\log_e x$ differentiates to $u'=\dfrac{1}{x}$, and $v=\dfrac{x^{2}}{2}$
   2. $\displaystyle \int \frac{1}{x} \cdot \frac{x^{2}}{2} \, dx$$\displaystyle = \frac{x^{2}}{4}$
3. 03Integrate$\displaystyle \int x \sin x \, dx = -x \cos x - (-\sin x)$AnswerHide

   Answer$\displaystyle -x \cos x + \sin x+C$
4. 04What is the $uv$ part?$\displaystyle \begin{gathered} \displaystyle \int x \log_e x \, dx \\[6pt] \displaystyle (u=\log_e x,\; v'=x \;\Rightarrow\; v=\frac{x^{2}}{2}) \end{gathered}$AnswerHide

   Answer$\displaystyle \frac{x^{2}}{2} \log_e x$

   Steps

   1. $v$ is the integral of $v'$; multiply it by $u$
   2. $\displaystyle (\log_e x) \cdot \frac{x^{2}}{2}$$\displaystyle = \frac{x^{2}}{2} \log_e x$
5. 05What is the $uv$ part?$\displaystyle \begin{gathered} \displaystyle \int x \sin x \, dx \\[6pt] \displaystyle (u=x,\; v'=\sin x \;\Rightarrow\; v=-\cos x) \end{gathered}$AnswerHide

   Answer$\displaystyle -x \cos x$

   Steps

   1. $v$ is the integral of $v'$; multiply it by $u$
   2. $\displaystyle x \cdot (-\cos x)$$\displaystyle = -x \cos x$
6. 06What is $\int u'v\,dx$? (omit $C$)$\displaystyle \int \log_e x \, dx = x \log_e x - \int u'v \, dx$AnswerHide

   Answer$\displaystyle x$

   Steps

   1. $u=\log_e x$ differentiates to $u'=\dfrac{1}{x}$, and $v=x$
   2. $\displaystyle \int \frac{1}{x} \cdot x \, dx$$\displaystyle = x$

FAQ