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

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

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integral of (4x+3)/(2x^2+3x)ln(2x^2+3x)

\int \frac{4 x + 3}{2 x ^{2} + 3 x} ln (2 x^{2} + 3 x) d x

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

Solution steps

\int \frac{4 x + 3}{2 x ^{2} + 3 x} ln (2 x^{2} + 3 x) d x

Apply u-substitution

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

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

Apply u-substitution

: \int v d v

= \int v d v

Apply the Power Rule

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

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

Substitute back

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

Add a constant to the solution

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

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