What is the integral of (9x+7)/((18x^2+28x)^4) ?

The integral of (9x+7)/((18x^2+28x)^4) is -1/(12(18x^2+28x)^3)+C

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integral of (9x+7)/((18x^2+28x)^4)

\int \frac{9 x + 7}{( 18 x ^{2} + 28 x ) ^{4}} d x

- \frac{1}{12 ( 18 x ^{2} + 28 x ) ^{3}} + C

Solution steps

\int \frac{9 x + 7}{( 18 x ^{2} + 28 x ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{4}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{4} (- \frac{1}{3 u ^{3}})

Substitute back

u = 18 x^{2} + 28 x

= \frac{1}{4} (- \frac{1}{3 ( 18 x ^{2} + 28 x ) ^{3}})

Simplify

\frac{1}{4} (- \frac{1}{3 ( 18 x ^{2} + 28 x ) ^{3}}) : - \frac{1}{12 ( 18 x ^{2} + 28 x ) ^{3}}

= - \frac{1}{12 ( 18 x ^{2} + 28 x ) ^{3}}

Add a constant to the solution

= - \frac{1}{12 ( 18 x ^{2} + 28 x ) ^{3}} + C

\int ln (x^{2} + 9) d x

\int_{- 1}^{1} (e^{y} - (y^{2} - 2)) d y

\int_{0}^{t} (t - x) sin (x) d x

y^{^{'}} + \frac{2}{x} y = 6 \frac{y ^{3}}{x ^{2}}

f (x) = 5 x (1 + x^{2})

What is the integral of (9x+7)/((18x^2+28x)^4) ?
The integral of (9x+7)/((18x^2+28x)^4) is -1/(12(18x^2+28x)^3)+C