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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}}

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

\int (x^{8} + sec^{2} (x)) d x

f (x) = \frac{( x ^{3} + 9 )}{x ^{3}}

x \to 0 lim (tan (3 x))

\int \frac{x - 2}{x} d x

18 t

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