What is the integral of (2x+3)/((2x^2+6x-1)^5) ?

The integral of (2x+3)/((2x^2+6x-1)^5) is -1/(8(2x^2+6x-1)^4)+C

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integral of (2x+3)/((2x^2+6x-1)^5)

\int \frac{2 x + 3}{( 2 x ^{2} + 6 x - 1 ) ^{5}} d x

- \frac{1}{8 ( 2 x ^{2} + 6 x - 1 ) ^{4}} + C

Solution steps

\int \frac{2 x + 3}{( 2 x ^{2} + 6 x - 1 ) ^{5}} d x

Apply u-substitution

: \int \frac{16 u}{( u ^{2} - 11 ) ^{5}} d u

= \int \frac{16 u}{( u ^{2} - 11 ) ^{5}} d u

Take the constant out:

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

= 16 \cdot \int \frac{u}{( u ^{2} - 11 ) ^{5}} d u

Apply u-substitution

: \int \frac{1}{2 ( v - 11 ) ^{5}} d v

= 16 \cdot \int \frac{1}{2 ( v - 11 ) ^{5}} d v

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int \frac{1}{( v - 11 ) ^{5}} d v

Apply u-substitution

: \int \frac{1}{w ^{5}} d w

= 16 \cdot \frac{1}{2} \cdot \int \frac{1}{w ^{5}} d w

Apply the Power Rule

: - \frac{1}{4 w ^{4}}

= 16 \cdot \frac{1}{2} (- \frac{1}{4 w ^{4}})

Substitute back

= 16 \cdot \frac{1}{2} - \frac{1}{4 ( ( 2 x + 3 ) ^{2} - 11 ) ^{4}}

Simplify

16 \cdot \frac{1}{2} - \frac{1}{4 ( ( 2 x + 3 ) ^{2} - 11 ) ^{4}} : - \frac{1}{8 ( 2 x ^{2} + 6 x - 1 ) ^{4}}

= - \frac{1}{8 ( 2 x ^{2} + 6 x - 1 ) ^{4}}

Add a constant to the solution

= - \frac{1}{8 ( 2 x ^{2} + 6 x - 1 ) ^{4}} + C

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What is the integral of (2x+3)/((2x^2+6x-1)^5) ?
The integral of (2x+3)/((2x^2+6x-1)^5) is -1/(8(2x^2+6x-1)^4)+C