What is the integral of 7x(2x+1)^{-3} ?

The integral of 7x(2x+1)^{-3} is (7(-4x-1))/(8(2x+1)^2)+C

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integral of 7x(2x+1)^{-3}

\int 7 x (2 x + 1)^{- 3} d x

\frac{7 ( - 4 x - 1 )}{8 ( 2 x + 1 ) ^{2}} + C

Solution steps

\int 7 x (2 x + 1)^{- 3} d x

Take the constant out:

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

= 7 \cdot \int x (2 x + 1)^{- 3} d x

Simplify

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

= 7 \cdot \int \frac{x}{( 2 x + 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{u - 1}{4 u ^{3}} d u

= 7 \cdot \int \frac{u - 1}{4 u ^{3}} d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int \frac{u - 1}{u ^{3}} d u

Expand

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

= 7 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{2}} - \frac{1}{u ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 \cdot \frac{1}{4} (\int \frac{1}{u ^{2}} d u - \int \frac{1}{u ^{3}} d u)

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

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

= 7 \cdot \frac{1}{4} (- \frac{1}{u} - (- \frac{1}{2 u ^{2}}))

Substitute back

u = 2 x + 1

= 7 \cdot \frac{1}{4} (- \frac{1}{2 x + 1} - (- \frac{1}{2 ( 2 x + 1 ) ^{2}}))

Simplify

7 \cdot \frac{1}{4} (- \frac{1}{2 x + 1} - (- \frac{1}{2 ( 2 x + 1 ) ^{2}})) : \frac{7 ( - 4 x - 1 )}{8 ( 2 x + 1 ) ^{2}}

= \frac{7 ( - 4 x - 1 )}{8 ( 2 x + 1 ) ^{2}}

Add a constant to the solution

= \frac{7 ( - 4 x - 1 )}{8 ( 2 x + 1 ) ^{2}} + C

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What is the integral of 7x(2x+1)^{-3} ?
The integral of 7x(2x+1)^{-3} is (7(-4x-1))/(8(2x+1)^2)+C