What is the integral from 0 to 1 of 1/((x^2+1)^3) ?

The integral from 0 to 1 of 1/((x^2+1)^3) is (3pi+8)/(32)

integral from 0 to 1 of 1/((x^2+1)^3)

\int_{0}^{1} \frac{1}{( x ^{2} + 1 ) ^{3}} d x

\frac{3 π + 8}{32}

Decimal

0.54452 \dots

Solution steps

\int_{0}^{1} \frac{1}{( x ^{2} + 1 ) ^{3}} d x

Compute the indefinite integral:

\int \frac{1}{( x ^{2} + 1 ) ^{3}} d x = \frac{3}{8} (arctan (x) + \frac{x}{1 + x ^{2}}) + \frac{x}{4 ( x ^{2} + 1 ) ^{2}} + C

Compute the boundaries:

\int_{0}^{1} \frac{1}{( x ^{2} + 1 ) ^{3}} d x = \frac{3 π + 8}{32} - 0

= \frac{3 π + 8}{32} - 0

Simplify

= \frac{3 π + 8}{32}

t an g e n t f (x) = 3 x^{2} - 5 x + 8, at x = 2

in v erse l a pl a ce \frac{2}{( s ^{2} - 1 ) ( s - 1 )}

\frac{\partial}{\partial y} (x^{2} + 5 y^{2})

x \to 3 - lim (- \frac{3}{7} x + \frac{4}{11})

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

What is the integral from 0 to 1 of 1/((x^2+1)^3) ?
The integral from 0 to 1 of 1/((x^2+1)^3) is (3pi+8)/(32)