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

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

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integral of (2x)/((x^2+4)ln(x^2+4))

\int \frac{2 x}{( x ^{2} + 4 ) ln ( x ^{2} + 4 )} d x

ln ln (x^{2} + 4) + C

Solution steps

\int \frac{2 x}{( x ^{2} + 4 ) ln ( x ^{2} + 4 )} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{( x ^{2} + 4 ) ln ( x ^{2} + 4 )} d x

Apply u-substitution

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

= 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

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

Apply u-substitution

: \int \frac{1}{v} d v

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

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

= ln ln (x^{2} + 4)

Add a constant to the solution

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

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