# Integrate by Parts, the IS-ID Way

In calculus, we learn to integrate products like $x \ln{x}$ and $e^x \sin{x}$ using the technique integration by parts. We use the formula,

$\displaystyle \int uv' \, dx = uv - \int u'v \, dx$,

and using LIATE (Log-InvTrig-Algebraic-Trig-Exp), choose the right-leaning term as $v'$ (e.g. $x$) and the left-leaning term  as $u$ (e.g. $\ln{x}$). We integrate and differentiate respectively to get $v=\displaystyle \frac{x^2}{2}$ and $u'=\displaystyle \frac{1}{x}$ respectively, substitute into the formula, simplify the expression and carry on our calculations. Rather than confuse you with which expression to use as $\displaystyle v'$ and $\displaystyle u$, I propose a simpler presentation.

Before we discover this presentation, I want to clarify that this is just a simpler presentation of the same technique. The concept is the same. The presentation simply makes more intuitive sense. Also, I might derive the formula in another post. Integration by parts is essentially what I call the “reverse product rule”, and I might elaborate more next time. For now, lets take a look at the formula one more time, but this time I highlighted in green $v$ and $v'$ and red $u$ and $u'$, and swapped the order of multiplying (doesn’t change the formula mathematically):

$\displaystyle \int$ $v' \,$ $u$ $dx =$ $v$ $u$  $- \displaystyle \int$ $v \,$ $u'$ $dx$.

Notice that $v' \,$ gets Integrated and $u$ remains the Same in the first term, $v$ $u$. Notice that $v' \,$ gets Integrated and $u$ gets Differentiated in the second term, $\displaystyle \int$ $v \,$ $u'$ $dx$. Thus, we get this idea of Integrate * Same – Integrate * Differentiate, or succinctly put,

$I$ $S$  $- \displaystyle \int$ $I \,$ $D$ $dx$.

As mentioned, this is essentially the same method, but using these letters makes it easier for us to choose which terms to use. Let’s integrate $x \ln{x}$ as an example.

Between $x$ and $\ln{x}$, the right-leaning term is $x$. Thus, we Integrate that to get $\displaystyle \frac{x^2}{2}$. The left-leaning term is $\ln{x}$. Thus, we Differentiate that to get $\displaystyle \frac{1}{x}$.

Using IS-ID, we plug in the relevant letters to get

$\displaystyle \int$ $x \,$ $\ln{x}$ $dx =$ $\displaystyle \frac{x^2}{2}$ $\ln{x}$  $- \displaystyle \int$ $\displaystyle \frac{x^2}{2}$ $\displaystyle \frac{1}{x}$ $dx$ $\displaystyle = \frac{x^2}{2}\ln{x} - \frac{1}{2}\int x \, dx$.

Integrating the remaining portion, we get our final result,

$\displaystyle \int x \ln{x}\, dx = \frac{x^2}{2}\ln{x} - \frac{x^2}{4} + C$.

I hope this alternative presentation will help you more effectively and integrate by parts with better understanding. As an exercise to the reader, find $\displaystyle \int e^x \sin{x}\, dx$. The final answer, for your reference, is $\displaystyle \frac{e^x}{2} (\sin{x} - \cos{x})$ and you get bonus marks if you can explain how this answer can be improved.