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

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

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

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

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

Solution steps

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

Simplify

ln ((x^{2} + 1)^{2}) : 2 ln (x^{2} + 1)

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

Take the constant out:

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

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

Apply u-substitution

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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