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

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

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

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

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

Solution steps

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

Factor

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

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

= 2 (- \frac{1}{3 u ^{3}})

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int x^{2} (x - 1)^{4} d x

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\int \frac{t ^{2}}{( 1 + t ^{2} ) ^{2}} d t

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