What is the integral from 1 to infinity of y/(y^4+1) ?

The integral from 1 to infinity of y/(y^4+1) is pi/8

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integral from 1 to infinity of y/(y^4+1)

\int_{1}^{\infty} \frac{y}{y ^{4} + 1} d y

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

\int_{1}^{\infty} \frac{y}{y ^{4} + 1} d y

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{\infty} \frac{y}{y ^{4} + 1} d y = \frac{π}{4} - \frac{π}{8}

= \frac{π}{4} - \frac{π}{8}

Simplify

= \frac{π}{8}

e^{9 x}

f (t) = sin^{2} (t)

\frac{x ^{3}}{2}

\int 8 d x

\int 50 x^{9} e^{5 x^{5}} d x

What is the integral from 1 to infinity of y/(y^4+1) ?
The integral from 1 to infinity of y/(y^4+1) is pi/8