Integration by parts

$$ \int u \, dv = uv = \int v\, du $$

or:

$$ \int_a^b u(x)v'(x)\,dx = [u(x)v(x)]^b_a - \int_a^b u'(x) v(x)\,dx \\ = u(b)v(b) - u(a)v(a) - \int_a^b u'(x)v(x)\,dx $$

Example: we want:

$$ I = \int \frac{\ln(x)}{x^2} \, dx $$

We note that:

$$ \frac{d}{dx} \ln(x) = \frac{1}{x} $$

We can write $I$ as:

$$ I = \int \ln(x) \frac{1}{x^2}\,dx $$

$$ = \int u(x) v'(x) \, dx $$

Where:

Therefore:

$$ I = \int u(x)v'(x)\, dx = uv - \int u'(x) u(x)\, dx $$

$$ = -\frac{\ln(x)}{x} - \int \frac{1}{x} (-\frac{1}{x})\,dx $$

$$ = -\frac{\ln(x)}{a} + \int x^{-2}\,dx $$

$$ = -\frac{ln(x)}{x} + (-x^{-1}) $$

$$ = -\frac{\ln(x)}{x} - \frac{1}{x} $$

$$ = - \frac{\ln(x) + 1}{x} $$