What is the integral of 3x(1+x)^{-4} ?

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

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integral of 3x(1+x)^{-4}

integral

3 x (1 + x)^{- 4}

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

Solution steps

\int 3 x (1 + x)^{- 4} d x

Take the constant out:

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

= 3 \cdot \int x (1 + x)^{- 4} d x

Apply u-substitution

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

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

Expand

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

= 3 \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

= 3 (\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}}

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

Substitute back

u = 1 + x

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

Simplify

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

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

Add a constant to the solution

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

\frac{u ^{2}}{( u ^{2} + 1 )}

cos^{4} (x) sin^{5} (x)

3^{x + 2}

\frac{1}{( a ^{2} + x ^{2} ) ^{0.5}}

\frac{sin ^{2} ( x )}{( 1 + sin ( x ) )}

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