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

The integral of (t^2+1)^2 is (t^5)/5+(2t^3)/3+t+C

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integral of (t^2+1)^2

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

\frac{t ^{5}}{5} + \frac{2 t ^{3}}{3} + t + C

Solution steps

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

Expand

(t^{2} + 1)^{2} : t^{4} + 2 t^{2} + 1

= \int t^{4} + 2 t^{2} + 1 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^{4} d t + \int 2 t^{2} d t + \int 1 d t

\int t^{4} d t = \frac{t ^{5}}{5}

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

\int 1 d t = t

= \frac{t ^{5}}{5} + \frac{2 t ^{3}}{3} + t

Add a constant to the solution

= \frac{t ^{5}}{5} + \frac{2 t ^{3}}{3} + t + C

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What is the integral of (t^2+1)^2 ?
The integral of (t^2+1)^2 is (t^5)/5+(2t^3)/3+t+C