What is the integral from 0 to 1 of 1/((x+1)(x^2+1)) ?

The integral from 0 to 1 of 1/((x+1)(x^2+1)) is (2ln(2)+pi)/8

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integral from 0 to 1 of 1/((x+1)(x^2+1))

\int_{0}^{1} \frac{1}{( x + 1 ) ( x ^{2} + 1 )} d x

\frac{2 ln ( 2 ) + π}{8}

Decimal

0.56598 \dots

Solution steps

\int_{0}^{1} \frac{1}{( x + 1 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

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

= \int_{0}^{1} \frac{1}{2 ( x + 1 )} + \frac{- x + 1}{2 ( x ^{2} + 1 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} \frac{1}{2 ( x + 1 )} d x + \int_{0}^{1} \frac{- x + 1}{2 ( x ^{2} + 1 )} d x

\int_{0}^{1} \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln (2)

\int_{0}^{1} \frac{- x + 1}{2 ( x ^{2} + 1 )} d x = \frac{1}{2} (- \frac{1}{2} ln (2) + \frac{π}{4})

= \frac{1}{2} ln (2) + \frac{1}{2} (- \frac{1}{2} ln (2) + \frac{π}{4})

Simplify

\frac{1}{2} ln (2) + \frac{1}{2} (- \frac{1}{2} ln (2) + \frac{π}{4}) : \frac{2 ln ( 2 ) + π}{8}

= \frac{2 ln ( 2 ) + π}{8}

f (x) = x^{4} + 8 x^{3} + 2 x + 2, at x = 1

\frac{\partial}{\partial x} (\frac{12 x}{x + y})

f (x) = (- x^{3} + x^{2} + x) cos (x)

\int x^{2} - 2 x + 1 d x

\int x^{2 k} d x

What is the integral from 0 to 1 of 1/((x+1)(x^2+1)) ?
The integral from 0 to 1 of 1/((x+1)(x^2+1)) is (2ln(2)+pi)/8