Question

integration x. log(x+1) dx .
explain in detail.​

Answer 1

QUESTION :-

$\rm{integration \:x.\: log(x+1) \:dx .}$

SOLUTION:-

Use LIATE.(logarithm, inverse, algebraic,tragnometry,exponential ) Accordingly decide u and v. in our case x. log(x+1) so by LIATE take u = log(x+1) and v = x So, integration by parts,

$\rm \implies \int vu \: dx = u \int \: vdx - \: \int[ \frac{du}{dx} \: \int \: vdx]dx \\ \rm \implies \: \int \: log(x + 1)xdx = log(x + 1) \int \: xdx - \int \: [ \frac{dlog(x + 1)}{dx} \: \int \: xdx]dx \: \\ \rm \implies \: log(x + 1) \frac{x {}^{2} }{2} - \frac{1}{2} - \int \: \frac{1}{x + 1} . \frac{x {}^{2} }{ 2} \: dx \\ \rm \: \green \: \tiny{[now \: \: x + 1 \sqrt{x {}^{2} } , i.e \: divide \: x {}^{2} \: by \: x + 1 \: ]} \\ \rm \: \: \tiny[ \: we \: get \: x {}^{2} = (x - 1)(x + 1) + 1] \: \\ \rm \implies \: log(x + 1) \: \frac{x {}^{2} }{2} \ - \frac{1}{2} \int \frac{(x - 1)(x + 1) + 1}{(x + 1)} .dx \\ \rm \implies \: log(x + 1) \: \frac{x {}^{2} }{2} - \frac{1}{2} \: \int \: \frac{(x - 1) \: (x + 1)}{(x + 1)} + \frac{1}{x + 1} .dx \\ \rm \implies \: log(x + 1) \: \frac{x {}^{2} }{2} - \frac{1}{2} \: \int \: \: x - 1 \: + \: \frac{1}{x + 1} .dx \: \\\rm \implies \: log(x + 1) \: \frac{x {}^{2} }{2} - \frac{1}{2} \: [ \frac{x {}^{2} }{2 } - x + log |x + 1| ] + c \: \\ \rm\implies { \boxed{log(x + 1) \: \frac{x {}^{2} }{2} - \frac{x {}^{2} }{4} + \: \frac{x}{2} - \frac{1}{2} \: log |x + 1| + c }}$