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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{2} d u

= \int \frac{u}{2} 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 u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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