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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = t^{3}

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

Add a constant to the solution

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

\frac{w}{w + \frac{c}{w}}

y = x^{2} - 6, at x = a

\frac{x ^{2} - 4}{x - 3}

\frac{d}{d θ} (1 - 2 sin (θ))

x 7^{x}

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