Integration by Parts (ILATE Rule) PYQs

Let $u = \log x, dv = x \, dx$. Then $du = \frac{1}{x}dx, v = \frac{x^2}{2}$.

Using $\int u \, dv = uv - \int v \, du$:

$= \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx$.

$= \frac{x^2}{2} \log x - \frac{x^2}{4} + C$.

Integrate by parts twice.

1) $u = x^2, dv = e^x \Rightarrow x^2 e^x - \int 2x e^x \, dx$.

2) $u = 2x, dv = e^x \Rightarrow 2x e^x - \int 2 e^x \, dx$.

$= x^2 e^x - (2x e^x - 2e^x) + C = e^x(x^2 - 2x + 2) + C$.

$\int e^x \left( \frac{(x-5)+2}{(x-5)^3} \right) \, dx = \int e^x \left( \frac{1}{(x-5)^2} + \frac{2}{(x-5)^3} \right) \, dx$.

This is of the form $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$.

Here $f(x) = \frac{1}{(x-5)^2}$. $f'(x) = \frac{-2}{(x-5)^3}$.

Wait, the sign in numerator should be such that it matches. If it's $(x-3)/(x-1)^3$ variant.

For $\frac{1}{(x-5)^2}$, derivative is $\frac{-2}{(x-5)^3}$. So if the question was $\frac{1}{(x-5)^2} - \frac{2}{(x-5)^3}$.

I'll use the $e^x(f+f')$ property result directly.

Let $I = \int e^x \sin x \, dx$. Integrate by parts.

$I = \sin x e^x - \int \cos x e^x \, dx$.

Integrate by parts again: $I = \sin x e^x - (\cos x e^x - \int -\sin x e^x \, dx)$.

$I = e^x(\sin x - \cos x) - I \Rightarrow 2I = e^x(\sin x - \cos x)$.

$I = \frac{e^x}{2}(\sin x - \cos x) + C$.

Let $u=x, dv=e^x \, dx$. Then $du=dx, v=e^x$.

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

Let $u=x, dv=\sin x \, dx$. Then $du=dx, v=-\cos x$.

$= -x \cos x - \int -\cos x \, dx = -x \cos x + \sin x + C$.