What is the integral of (t^2+1)^5t ?

The integral of (t^2+1)^5t is 1/12 (t^2+1)^6+C

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integral of (t^2+1)^5t

\int (t^{2} + 1)^{5} t d t

\frac{1}{12} (t^{2} + 1)^{6} + C

Solution steps

\int (t^{2} + 1)^{5} t d t

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = t^{2} + 1

= \frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{6}}{6} : \frac{1}{12} (t^{2} + 1)^{6}

= \frac{1}{12} (t^{2} + 1)^{6}

Add a constant to the solution

= \frac{1}{12} (t^{2} + 1)^{6} + C

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What is the integral of (t^2+1)^5t ?
The integral of (t^2+1)^5t is 1/12 (t^2+1)^6+C