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

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

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integral of t/((t+1)^4)

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

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

Solution steps

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

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 = t + 1

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

Simplify

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

Add a constant to the solution

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

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f (x) = 10

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