What is the integral from 1 to 2 of 1/((x+1)^2) ?

The integral from 1 to 2 of 1/((x+1)^2) is 1/6

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integral from 1 to 2 of 1/((x+1)^2)

\int_{1}^{2} \frac{1}{( x + 1 ) ^{2}} d x

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

\int_{1}^{2} \frac{1}{( x + 1 ) ^{2}} d x

Apply u-substitution

: \int_{2}^{3} \frac{1}{u ^{2}} d u

= \int_{2}^{3} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{2}^{3}

= [- \frac{1}{u}]_{2}^{3}

Compute the boundaries:

\frac{1}{6}

= \frac{1}{6}

f (x) = ∣ x - 1 ∣

\frac{\partial}{\partial y} (ln (x + 3 y))

\int \frac{x - 2}{x ^{2} - x} d x

\int_{0}^{2} 2^{x} d x

\int \frac{1}{3 x ^{2} + 1} d x

What is the integral from 1 to 2 of 1/((x+1)^2) ?
The integral from 1 to 2 of 1/((x+1)^2) is 1/6