Question

Please give me solution of this integral.

Answer should be log(x²+1)+1/x²+1 + C

Don't write fake answers...​

Answer 1

Hope this'll help you.

Answer 2

$\rm \implies \displaystyle \int \rm \dfrac{2 {x}^{3} }{ {( {x}^{2} + 1) }^{2} } dx \\ \\ \\ \rm let \: \: \: {x}^{2} + 1 = t \\ \rm \: 2x \: dx = dt \\ \\ \boxed{ {x}^{2} = t - 1 } \\ \\ \\ \rm \implies \displaystyle \int \rm \dfrac{{x}^{2} \: 2x \: dx}{ {( {x}^{2} + 1) }^{2} }\\ \\ \\ \rm \implies \displaystyle \int \rm \dfrac{{(t - 1)} \: dt}{ { t }^{2} }\\ \\ \\ \rm \implies \displaystyle \int \rm \dfrac{{(t )} \: dt}{ { t }^{2} } - \displaystyle \int \rm \dfrac{{ } \: dt}{ { t }^{2}}\\ \\ \\ \rm \implies \displaystyle \int \rm \dfrac{ dt}{ { t } } - \displaystyle \int \rm \dfrac{ \: dt}{ { t }^{2}}\\ \\ \\ \rm \implies \displaystyle \rm log \: t \rm - \displaystyle \int \rm { t }^{ - 2}dt\\ \\ \\ \rm \implies \displaystyle \rm log \: t \rm - ( \rm { t }^{ - 2 + 1} \times \frac{1}{ - 2 + 1} ) + c\\ \\ \\ \rm \implies \displaystyle \rm log \: t \rm - ( \rm { t }^{ - 2 + 1} \times \frac{1}{ - 1} ) + c\\ \\ \\ \rm \implies \displaystyle \rm log \: t \rm \: + \: \rm { t }^{ - 1} + c \\ \\ \\ \rm \implies \displaystyle \rm log \: t \rm \: + \dfrac{1}{t} + c\\ \\ \\ \rm \implies \displaystyle \rm log \: ( {x}^{2} + 1) \rm \: + \dfrac{1}{( {x}^{2} + 1) } + c$