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

The integral of (xln^2(1+x^2))/(1+x^2) is 1/6 ln^3(1+x^2)+C

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

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

\frac{1}{6} ln^{3} (1 + x^{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ^{2} ( u )}{2 u} d u

= \int \frac{ln ^{2} ( u )}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{ln ^{2} ( u )}{u} d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{2} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{2} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{2} \cdot \frac{ln ^{3} ( 1 + x ^{2} )}{3}

Simplify

\frac{1}{2} \cdot \frac{ln ^{3} ( 1 + x ^{2} )}{3} : \frac{1}{6} ln^{3} (1 + x^{2})

= \frac{1}{6} ln^{3} (1 + x^{2})

Add a constant to the solution

= \frac{1}{6} ln^{3} (1 + x^{2}) + C

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What is the integral of (xln^2(1+x^2))/(1+x^2) ?
The integral of (xln^2(1+x^2))/(1+x^2) is 1/6 ln^3(1+x^2)+C