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

The integral of sqrt(t)(t^2+t-1) is sqrt(t)((t^3)/3+(t^2)/2-t)-1/210 t^{3/2}(10t^2+21t-70)+C

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

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

t (\frac{t ^{3}}{3} + \frac{t ^{2}}{2} - t) - \frac{1}{210} t^{\frac{3}{2}} (10 t^{2} + 21 t - 70) + C

Solution steps

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

Apply Integration By Parts

: t (\frac{t ^{3}}{3} + \frac{t ^{2}}{2} - t) - \int \frac{1}{12} t (2 t^{2} + 3 t - 6) d t

= t (\frac{t ^{3}}{3} + \frac{t ^{2}}{2} - t) - \int \frac{1}{12} t (2 t^{2} + 3 t - 6) d t

\int \frac{1}{12} t (2 t^{2} + 3 t - 6) d t = \frac{1}{210} t^{\frac{3}{2}} (10 t^{2} + 21 t - 70)

= t (\frac{t ^{3}}{3} + \frac{t ^{2}}{2} - t) - \frac{1}{210} t^{\frac{3}{2}} (10 t^{2} + 21 t - 70)

Add a constant to the solution

= t (\frac{t ^{3}}{3} + \frac{t ^{2}}{2} - t) - \frac{1}{210} t^{\frac{3}{2}} (10 t^{2} + 21 t - 70) + C

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What is the integral of sqrt(t)(t^2+t-1) ?
The integral of sqrt(t)(t^2+t-1) is sqrt(t)((t^3)/3+(t^2)/2-t)-1/210 t^{3/2}(10t^2+21t-70)+C