What is the integral of (ln((3+x^2)^x))/(3+x^2) ?

The integral of (ln((3+x^2)^x))/(3+x^2) is 1/4 ln^2(3+x^2)+C

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integral of (ln((3+x^2)^x))/(3+x^2)

\int \frac{ln ( ( 3 + x ^{2} ) ^{x} )}{3 + x ^{2}} d x

\frac{1}{4} ln^{2} (3 + x^{2}) + C

Solution steps

\int \frac{ln ( ( 3 + x ^{2} ) ^{x} )}{3 + x ^{2}} d x

Apply u-substitution

: \int \frac{ln ( 3 + u )}{2 ( 3 + u )} d u

= \int \frac{ln ( 3 + u )}{2 ( 3 + 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 ( 3 + u )}{3 + u} d u

Apply u-substitution

: \int \frac{ln ( v )}{v} d v

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

Apply u-substitution

: \int w d w

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

Apply the Power Rule

: \frac{w ^{2}}{2}

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( 3 + x ^{2} )}{2} : \frac{1}{4} ln^{2} (3 + x^{2})

= \frac{1}{4} ln^{2} (3 + x^{2})

Add a constant to the solution

= \frac{1}{4} ln^{2} (3 + x^{2}) + C

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What is the integral of (ln((3+x^2)^x))/(3+x^2) ?
The integral of (ln((3+x^2)^x))/(3+x^2) is 1/4 ln^2(3+x^2)+C