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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

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

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

v = 1 + u

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

Add a constant to the solution

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

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