What is the integral of (48)/((t+1)^4)-3 ?

The integral of (48)/((t+1)^4)-3 is -(16)/((t+1)^3)-3t+C

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integral of (48)/((t+1)^4)-3

\int \frac{48}{( t + 1 ) ^{4}} - 3 d t

- \frac{16}{( t + 1 ) ^{3}} - 3 t + C

Solution steps

\int \frac{48}{( t + 1 ) ^{4}} - 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{48}{( t + 1 ) ^{4}} d t - \int 3 d t

\int \frac{48}{( t + 1 ) ^{4}} d t = - \frac{16}{( t + 1 ) ^{3}}

\int 3 d t = 3 t

= - \frac{16}{( t + 1 ) ^{3}} - 3 t

Add a constant to the solution

= - \frac{16}{( t + 1 ) ^{3}} - 3 t + C

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What is the integral of (48)/((t+1)^4)-3 ?
The integral of (48)/((t+1)^4)-3 is -(16)/((t+1)^3)-3t+C