What is the integral of (xln(x))/((1+x^2)^2) ?

The integral of (xln(x))/((1+x^2)^2) is -(ln(x))/(2(1+x^2))+1/2 (ln|x|-1/2 ln|x^2+1|)+C

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integral of (xln(x))/((1+x^2)^2)

\int \frac{x ln ( x )}{( 1 + x ^{2} ) ^{2}} d x

- \frac{ln ( x )}{2 ( 1 + x ^{2} )} + \frac{1}{2} (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) + C

Solution steps

\int \frac{x ln ( x )}{( 1 + x ^{2} ) ^{2}} d x

\frac{x ln ( x )}{( 1 + x ^{2} ) ^{2}} = x ln (x) \frac{1}{( 1 + x ^{2} ) ^{2}}

= \int x ln (x) \frac{1}{( 1 + x ^{2} ) ^{2}} d x

Apply Integration By Parts

: - \frac{ln ( x )}{2 ( 1 + x ^{2} )} - \int - \frac{1}{2 x ( 1 + x ^{2} )} d x

= - \frac{ln ( x )}{2 ( 1 + x ^{2} )} - \int - \frac{1}{2 x ( 1 + x ^{2} )} d x

\int - \frac{1}{2 x ( 1 + x ^{2} )} d x = - \frac{1}{2} (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1)

= - \frac{ln ( x )}{2 ( 1 + x ^{2} )} - (- \frac{1}{2} (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1))

Simplify

= - \frac{ln ( x )}{2 ( 1 + x ^{2} )} + \frac{1}{2} (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1)

Add a constant to the solution

= - \frac{ln ( x )}{2 ( 1 + x ^{2} )} + \frac{1}{2} (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) + C

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What is the integral of (xln(x))/((1+x^2)^2) ?
The integral of (xln(x))/((1+x^2)^2) is -(ln(x))/(2(1+x^2))+1/2 (ln|x|-1/2 ln|x^2+1|)+C