What is the integral of t^{7/2}(t^2-4t+17) ?

The integral of t^{7/2}(t^2-4t+17) is 2/13 t^{13/2}-8/11 t^{11/2}+34/9 t^{9/2}+C

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integral of t^{7/2}(t^2-4t+17)

\int t^{\frac{7}{2}} (t^{2} - 4 t + 17) d t

\frac{2}{13} t^{\frac{13}{2}} - \frac{8}{11} t^{\frac{11}{2}} + \frac{34}{9} t^{\frac{9}{2}} + C

Solution steps

\int t^{\frac{7}{2}} (t^{2} - 4 t + 17) d t

Expand

t^{\frac{7}{2}} (t^{2} - 4 t + 17) : t^{\frac{11}{2}} - 4 t^{\frac{9}{2}} + 17 t^{\frac{7}{2}}

= \int t^{\frac{11}{2}} - 4 t^{\frac{9}{2}} + 17 t^{\frac{7}{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{11}{2}} d t - \int 4 t^{\frac{9}{2}} d t + \int 17 t^{\frac{7}{2}} d t

\int t^{\frac{11}{2}} d t = \frac{2}{13} t^{\frac{13}{2}}

\int 4 t^{\frac{9}{2}} d t = \frac{8}{11} t^{\frac{11}{2}}

\int 17 t^{\frac{7}{2}} d t = \frac{34}{9} t^{\frac{9}{2}}

= \frac{2}{13} t^{\frac{13}{2}} - \frac{8}{11} t^{\frac{11}{2}} + \frac{34}{9} t^{\frac{9}{2}}

Add a constant to the solution

= \frac{2}{13} t^{\frac{13}{2}} - \frac{8}{11} t^{\frac{11}{2}} + \frac{34}{9} t^{\frac{9}{2}} + C

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What is the integral of t^{7/2}(t^2-4t+17) ?
The integral of t^{7/2}(t^2-4t+17) is 2/13 t^{13/2}-8/11 t^{11/2}+34/9 t^{9/2}+C