What is the integral of (3x^2+3x)/(2x^3+3x^2-5) ?

The integral of (3x^2+3x)/(2x^3+3x^2-5) is 1/2 ln|2x^3+3x^2-5|+C

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integral of (3x^2+3x)/(2x^3+3x^2-5)

\int \frac{3 x ^{2} + 3 x}{2 x ^{3} + 3 x ^{2} - 5} d x

\frac{1}{2} ln 2 x^{3} + 3 x^{2} - 5 + C

Solution steps

\int \frac{3 x ^{2} + 3 x}{2 x ^{3} + 3 x ^{2} - 5} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 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{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x^{3} + 3 x^{2} - 5

= \frac{1}{2} ln 2 x^{3} + 3 x^{2} - 5

Add a constant to the solution

= \frac{1}{2} ln 2 x^{3} + 3 x^{2} - 5 + C

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What is the integral of (3x^2+3x)/(2x^3+3x^2-5) ?
The integral of (3x^2+3x)/(2x^3+3x^2-5) is 1/2 ln|2x^3+3x^2-5|+C