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

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

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

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

\frac{π + 2}{8}

Decimal Notation

0.64269 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

= \int_{0}^{\frac{π}{4}} \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 1 + cos (2 u) d u

Apply the Sum Rule:

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

= \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d u + \int_{0}^{\frac{π}{4}} cos (2 u) d u)

\int_{0}^{\frac{π}{4}} 1 d u = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 u) d u = \frac{1}{2}

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

Simplify

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

= \frac{π + 2}{8}

\frac{d}{d x} (\frac{1 - 2 x ^{3}}{( 1 + x ^{3} ) ^{2}})

(arctan (8 x))^{2}

\int_{1}^{e^{2}} ln (x) d x

y^{^{''}} + 14 y^{^{'}} + 58 y = 0

\frac{1}{2 x} + \frac{1}{7 x ^{\frac{6}{7}}}

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