What is the integral from 0 to 1 of 4/((x+1)(x^2+1)) ?

The integral from 0 to 1 of 4/((x+1)(x^2+1)) is (2ln(2)+pi)/2

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integral from 0 to 1 of 4/((x+1)(x^2+1))

\int_{0}^{1} \frac{4}{( x + 1 ) ( x ^{2} + 1 )} d x

\frac{2 ln ( 2 ) + π}{2}

Decimal

2.26394 \dots

Solution steps

\int_{0}^{1} \frac{4}{( x + 1 ) ( x ^{2} + 1 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \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 )}

= 4 \cdot \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

= 4 (\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})

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

Simplify

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

= \frac{2 ln ( 2 ) + π}{2}

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What is the integral from 0 to 1 of 4/((x+1)(x^2+1)) ?
The integral from 0 to 1 of 4/((x+1)(x^2+1)) is (2ln(2)+pi)/2