What is the integral of (13t^7+t^5)/(t^{3/2)} ?

The integral of (13t^7+t^5)/(t^{3/2)} is 2t^{13/2}+2/9 t^{9/2}+C

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integral of (13t^7+t^5)/(t^{3/2)}

\int \frac{13 t ^{7} + t ^{5}}{t ^{\frac{3}{2}}} d t

2 t^{\frac{13}{2}} + \frac{2}{9} t^{\frac{9}{2}} + C

Solution steps

\int \frac{13 t ^{7} + t ^{5}}{t ^{\frac{3}{2}}} d t

Expand

\frac{13 t ^{7} + t ^{5}}{t ^{\frac{3}{2}}} : \frac{13 t ^{7}}{t ^{\frac{3}{2}}} + \frac{t ^{5}}{t ^{\frac{3}{2}}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{13 t ^{7}}{t ^{\frac{3}{2}}} d t + \int \frac{t ^{5}}{t ^{\frac{3}{2}}} d t

\int \frac{13 t ^{7}}{t ^{\frac{3}{2}}} d t = 2 t^{\frac{13}{2}}

\int \frac{t ^{5}}{t ^{\frac{3}{2}}} d t = \frac{2}{9} t^{\frac{9}{2}}

= 2 t^{\frac{13}{2}} + \frac{2}{9} t^{\frac{9}{2}}

Add a constant to the solution

= 2 t^{\frac{13}{2}} + \frac{2}{9} t^{\frac{9}{2}} + C

x \to 0 lim (x^{x^{2}})

y^{^{'}} + 0.15 y = 1

(2 x^{2} + y) d x + (x^{2} y - x) d y = 0

\frac{d y}{d x} = \frac{9 x}{5 y}

\int \frac{1}{2 x ^{2} + 8} d x

What is the integral of (13t^7+t^5)/(t^{3/2)} ?
The integral of (13t^7+t^5)/(t^{3/2)} is 2t^{13/2}+2/9 t^{9/2}+C