What is the integral from 0 to 1 of (t^2)/(1+t^6) ?

The integral from 0 to 1 of (t^2)/(1+t^6) is pi/(12)

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integral from 0 to 1 of (t^2)/(1+t^6)

\int_{0}^{1} \frac{t ^{2}}{1 + t ^{6}} d t

\frac{π}{12}

Decimal Notation

0.26179 \dots

Solution steps

\int_{0}^{1} \frac{t ^{2}}{1 + t ^{6}} d t

Compute the indefinite integral:

\int \frac{t ^{2}}{1 + t ^{6}} d t = \frac{1}{3} arctan (t^{3}) + C

Compute the boundaries:

\int_{0}^{1} \frac{t ^{2}}{1 + t ^{6}} d t = \frac{π}{12} - 0

= \frac{π}{12} - 0

Simplify

= \frac{π}{12}

\frac{x}{x + 3}

(2 x - 1)^{(2)} ln (2 x - 1)

\frac{d}{d x} (arctan (x^{2} - 2) + C)

\int \frac{sec ^{2} ( x )}{( 1 + tan ( x ) ) ^{2}} d x

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

What is the integral from 0 to 1 of (t^2)/(1+t^6) ?
The integral from 0 to 1 of (t^2)/(1+t^6) is pi/(12)