What is the integral of (4x+6)/((x^2+3x+7)^3) ?

The integral of (4x+6)/((x^2+3x+7)^3) is -1/((x^2+3x+7)^2)+C

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integral of (4x+6)/((x^2+3x+7)^3)

\int \frac{4 x + 6}{( x ^{2} + 3 x + 7 ) ^{3}} d x

- \frac{1}{( x ^{2} + 3 x + 7 ) ^{2}} + C

Solution steps

\int \frac{4 x + 6}{( x ^{2} + 3 x + 7 ) ^{3}} d x

Apply u-substitution

= \int \frac{2}{u ^{3}} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= 2 (- \frac{1}{2 u ^{2}})

Substitute back

u = x^{2} + 3 x + 7

= 2 (- \frac{1}{2 ( x ^{2} + 3 x + 7 ) ^{2}})

Simplify

2 (- \frac{1}{2 ( x ^{2} + 3 x + 7 ) ^{2}}) : - \frac{1}{( x ^{2} + 3 x + 7 ) ^{2}}

= - \frac{1}{( x ^{2} + 3 x + 7 ) ^{2}}

Add a constant to the solution

= - \frac{1}{( x ^{2} + 3 x + 7 ) ^{2}} + C

What is the integral of (4x+6)/((x^2+3x+7)^3) ?
The integral of (4x+6)/((x^2+3x+7)^3) is -1/((x^2+3x+7)^2)+C