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

The integral of x/(1+x^2)ln(1+x^2) is 1/4 ln^2(1+x^2)+C

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

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

\frac{1}{4} ln^{2} (1 + x^{2}) + C

Solution steps

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Apply u-substitution

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

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

Apply u-substitution

: \int v d v

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

Apply the Power Rule

: \frac{v ^{2}}{2}

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( 1 + x ^{2} )}{2} : \frac{1}{4} ln^{2} (1 + x^{2})

= \frac{1}{4} ln^{2} (1 + x^{2})

Add a constant to the solution

= \frac{1}{4} ln^{2} (1 + x^{2}) + C

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\frac{d}{d x} (e^{x^{2} + c o s (x)})

99 - 80 e^{- 0.3}

What is the integral of x/(1+x^2)ln(1+x^2) ?
The integral of x/(1+x^2)ln(1+x^2) is 1/4 ln^2(1+x^2)+C