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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = v + 1

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

Add a constant to the solution

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

3 t^{2} + 2 t + 1

3 t^{2} + 2 t + 2

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

\frac{( a ^{2} - x ^{2} ) ^{3}}{2}

\frac{sinh ( x )}{( 1 + cosh ^{2} ( x ) )}