What is the integral of t^{3/2}(t^2-4t+17) ?

The integral of t^{3/2}(t^2-4t+17) is 2/9 t^{9/2}-8/7 t^{7/2}+34/5 t^{5/2}+C

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integral of t^{3/2}(t^2-4t+17)

\int t^{\frac{3}{2}} (t^{2} - 4 t + 17) d t

\frac{2}{9} t^{\frac{9}{2}} - \frac{8}{7} t^{\frac{7}{2}} + \frac{34}{5} t^{\frac{5}{2}} + C

Solution steps

\int t^{\frac{3}{2}} (t^{2} - 4 t + 17) d t

Expand

t^{\frac{3}{2}} (t^{2} - 4 t + 17) : t^{\frac{7}{2}} - 4 t^{\frac{5}{2}} + 17 t^{\frac{3}{2}}

= \int t^{\frac{7}{2}} - 4 t^{\frac{5}{2}} + 17 t^{\frac{3}{2}} 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{7}{2}} d t - \int 4 t^{\frac{5}{2}} d t + \int 17 t^{\frac{3}{2}} d t

\int t^{\frac{7}{2}} d t = \frac{2}{9} t^{\frac{9}{2}}

\int 4 t^{\frac{5}{2}} d t = \frac{8}{7} t^{\frac{7}{2}}

\int 17 t^{\frac{3}{2}} d t = \frac{34}{5} t^{\frac{5}{2}}

= \frac{2}{9} t^{\frac{9}{2}} - \frac{8}{7} t^{\frac{7}{2}} + \frac{34}{5} t^{\frac{5}{2}}

Add a constant to the solution

= \frac{2}{9} t^{\frac{9}{2}} - \frac{8}{7} t^{\frac{7}{2}} + \frac{34}{5} t^{\frac{5}{2}} + C

\int \frac{2}{x ^{2} - 9} d x

\int (x x - 4 x^{3} + 5) d x

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

\int in d x

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

What is the integral of t^{3/2}(t^2-4t+17) ?
The integral of t^{3/2}(t^2-4t+17) is 2/9 t^{9/2}-8/7 t^{7/2}+34/5 t^{5/2}+C