What is the integral of t(t^2+1)^7 ?

The integral of t(t^2+1)^7 is 1/16 (t^2+1)^8+C

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integral of t(t^2+1)^7

\int t (t^{2} + 1)^{7} d t

\frac{1}{16} (t^{2} + 1)^{8} + C

Solution steps

\int t (t^{2} + 1)^{7} d t

Apply u-substitution

: \int \frac{u ^{7}}{2} d u

= \int \frac{u ^{7}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{1}{2} \cdot \frac{u ^{8}}{8}

Substitute back

u = t^{2} + 1

= \frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{8}}{8}

Simplify

\frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{8}}{8} : \frac{1}{16} (t^{2} + 1)^{8}

= \frac{1}{16} (t^{2} + 1)^{8}

Add a constant to the solution

= \frac{1}{16} (t^{2} + 1)^{8} + C

\frac{d y}{d x} = e^{x}

\frac{d}{d x} (\frac{a x + b}{c})

\int tan^{2} (t) d tan (t)

\int 16 - 3 x^{2} d x

\frac{d}{d x} ((1 + 7 x^{2}) (x - x^{2}))

What is the integral of t(t^2+1)^7 ?
The integral of t(t^2+1)^7 is 1/16 (t^2+1)^8+C