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

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

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

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

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

Solution steps

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

n = 0 \sum \infty \frac{n - 1}{n ^{2}}

\int 1 + e^{t} d t

y^{^{''}} + 0.1 y^{^{'}} + 0.0004 y = 0

f (x) = \frac{x - 6}{x + 2}, at x = 2

8 x - 3 x

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