Q.$\dfrac{\log x}{(1+\log x)^{2}}$

Integration by Parts

The idea: reverse the product rule

Some integrands are a product of two very different functions — $x e^x$, $x\cos x$, $\log x$, $x\sin^{-1}x$ — where substitution gets you nowhere. Integration by parts is the tool for these. It comes straight from reversing the product rule for differentiation.

Starting from $\dfrac{d}{dx}(uv)=u\,v' + u'\,v$ and integrating both sides gives the working formula:

In words: integral of (first $\times$ derivative-of-second) $=$ first $\times$ integral-of-second $-$ integral of (derivative-of-first $\times$ integral-of-second).

Choosing $u$: the ILATE rule

The whole game is picking which factor is $u$ (to differentiate) and which is $dv$ (to integrate). Pick $u$ by ILATE — the first type that appears:

• Inverse trig ($\sin^{-1}x$), Logarithmic ($\log x$), Algebraic ($x^2$), Trigonometric ($\sin x$), Exponential ($e^x$).

Whatever comes first in ILATE becomes $u$; the rest is $dv$. This makes the new integral $\int v\,du$ simpler than the one you started with.

Worked idea

For $\int x e^x\,dx$: algebraic before exponential, so $u=x$, $dv=e^x dx$. Then $du=dx$, $v=e^x$:

$\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C.$ …