What is the integral of (y^2)/(y^6+1) ?

The integral of (y^2)/(y^6+1) is 1/3 arctan(y^3)+C

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integral of (y^2)/(y^6+1)

\int \frac{y ^{2}}{y ^{6} + 1} d y

\frac{1}{3} arctan (y^{3}) + C

Solution steps

\int \frac{y ^{2}}{y ^{6} + 1} d y

Apply u-substitution

: \int \frac{1}{3 ( u ^{2} + 1 )} d u

= \int \frac{1}{3 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{3} arctan (u)

Substitute back

u = y^{3}

= \frac{1}{3} arctan (y^{3})

Add a constant to the solution

= \frac{1}{3} arctan (y^{3}) + C

n = 1 \sum \infty 7 (- 4)^{n - 1}

\frac{1}{( x ^{2} + 1 )}

5 x, x^{2}

x \to 0 lim (\frac{x}{x + 2})

\int sec^{4} (x) tan^{8} (x) d x

What is the integral of (y^2)/(y^6+1) ?
The integral of (y^2)/(y^6+1) is 1/3 arctan(y^3)+C