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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{8 u}{( u ^{2} - 29 ) ^{2}} d u

= \int \frac{8 u}{( u ^{2} - 29 ) ^{2}} d u

Take the constant out:

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

= 8 \cdot \int \frac{u}{( u ^{2} - 29 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{2 v ^{2}} d v

= 8 \cdot \int \frac{1}{2 v ^{2}} d v

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 8 \cdot \frac{1}{2} (- \frac{1}{v})

Substitute back

= 8 \cdot \frac{1}{2} (- \frac{1}{( 2 x + 3 ) ^{2} - 29})

Simplify

8 \cdot \frac{1}{2} (- \frac{1}{( 2 x + 3 ) ^{2} - 29}) : - \frac{1}{x ^{2} + 3 x - 5}

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

Add a constant to the solution

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

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