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

The integral of (t-1)(2+t^2) is t^2-2t+(t^4)/2-t^3-(t^4)/4+(2t^3)/3+C

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

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

t^{2} - 2 t + \frac{t ^{4}}{2} - t^{3} - \frac{t ^{4}}{4} + \frac{2 t ^{3}}{3} + C

Solution steps

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

Apply Integration By Parts

: t^{2} - 2 t + \frac{t ^{4}}{2} - t^{3} - \int t^{3} - 2 t^{2} d t

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

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

= t^{2} - 2 t + \frac{t ^{4}}{2} - t^{3} - (\frac{t ^{4}}{4} - \frac{2 t ^{3}}{3})

Simplify

= t^{2} - 2 t + \frac{t ^{4}}{2} - t^{3} - \frac{t ^{4}}{4} + \frac{2 t ^{3}}{3}

Add a constant to the solution

= t^{2} - 2 t + \frac{t ^{4}}{2} - t^{3} - \frac{t ^{4}}{4} + \frac{2 t ^{3}}{3} + C

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What is the integral of (t-1)(2+t^2) ?
The integral of (t-1)(2+t^2) is t^2-2t+(t^4)/2-t^3-(t^4)/4+(2t^3)/3+C