What is the integral of 1/((v+1)^2) ?

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

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integral of 1/((v+1)^2)

integral

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

- \frac{1}{v + 1} + C

Solution steps

\int \frac{1}{( v + 1 ) ^{2}} d v

Apply u-substitution

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= - \frac{1}{u}

Substitute back

u = v + 1

= - \frac{1}{v + 1}

Add a constant to the solution

= - \frac{1}{v + 1} + C