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

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

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integral of (2t-1)/((t+1)^3)

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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