What is the integral of (6x)/((1+x)^4) ?

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

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integral of (6x)/((1+x)^4)

\int \frac{6 x}{( 1 + x ) ^{4}} d x

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

Solution steps

\int \frac{6 x}{( 1 + x ) ^{4}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{( 1 + x ) ^{4}} d x

Apply u-substitution

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

= 6 \cdot \int \frac{u - 1}{u ^{4}} d u

Expand

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

= 6 \cdot \int \frac{1}{u ^{3}} - \frac{1}{u ^{4}} 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

= 6 (\int \frac{1}{u ^{3}} d u - \int \frac{1}{u ^{4}} d u)

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

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

= 6 (- \frac{1}{2 u ^{2}} - (- \frac{1}{3 u ^{3}}))

Substitute back

u = 1 + x

= 6 (- \frac{1}{2 ( 1 + x ) ^{2}} - (- \frac{1}{3 ( 1 + x ) ^{3}}))

Simplify

= 6 (- \frac{1}{2 ( 1 + x ) ^{2}} + \frac{1}{3 ( 1 + x ) ^{3}})

Add a constant to the solution

= 6 (- \frac{1}{2 ( 1 + x ) ^{2}} + \frac{1}{3 ( 1 + x ) ^{3}}) + C

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