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

The integral of t(t^2+1)^2 is (t^6)/6+(t^4)/2+(t^2)/2+C

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

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

\frac{t ^{6}}{6} + \frac{t ^{4}}{2} + \frac{t ^{2}}{2} + C

Solution steps

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

Expand

t (t^{2} + 1)^{2} : t^{5} + 2 t^{3} + t

= \int t^{5} + 2 t^{3} + t d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t^{5} d t + \int 2 t^{3} d t + \int t d t

\int t^{5} d t = \frac{t ^{6}}{6}

\int 2 t^{3} d t = \frac{t ^{4}}{2}

\int t d t = \frac{t ^{2}}{2}

= \frac{t ^{6}}{6} + \frac{t ^{4}}{2} + \frac{t ^{2}}{2}

Add a constant to the solution

= \frac{t ^{6}}{6} + \frac{t ^{4}}{2} + \frac{t ^{2}}{2} + C

What is the integral of t(t^2+1)^2 ?
The integral of t(t^2+1)^2 is (t^6)/6+(t^4)/2+(t^2)/2+C