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

The integral of sqrt(t)(t^2-1) is 2/7 t^{7/2}-2/3 t^{3/2}+C

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

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

\frac{2}{7} t^{\frac{7}{2}} - \frac{2}{3} t^{\frac{3}{2}} + C

Solution steps

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

Expand

t (t^{2} - 1) : t^{\frac{5}{2}} - t

= \int t^{\frac{5}{2}} - 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^{\frac{5}{2}} d t - \int t d t

\int t^{\frac{5}{2}} d t = \frac{2}{7} t^{\frac{7}{2}}

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

= \frac{2}{7} t^{\frac{7}{2}} - \frac{2}{3} t^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} t^{\frac{7}{2}} - \frac{2}{3} t^{\frac{3}{2}} + C

\frac{d}{d x} (3 e^{(x^{2})})

\int cos (x^{3}) d x

y = 7 x + 11, (2, 5)

e^{3 a - b t}

\int \frac{11 x ^{3}}{4 + x ^{2}} d x

What is the integral of sqrt(t)(t^2-1) ?
The integral of sqrt(t)(t^2-1) is 2/7 t^{7/2}-2/3 t^{3/2}+C