What is the integral of (sqrt(t)(2t^2-1)^2)/(t^3) ?

The integral of (sqrt(t)(2t^2-1)^2)/(t^3) is 8/5 t^{5/2}-8sqrt(t)-2/(3t^{3/2)}+C

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integral of (sqrt(t)(2t^2-1)^2)/(t^3)

\int \frac{t ( 2 t ^{2} - 1 ) ^{2}}{t ^{3}} d t

\frac{8}{5} t^{\frac{5}{2}} - 8 t - \frac{2}{3 t ^{\frac{3}{2}}} + C

Solution steps

\int \frac{t ( 2 t ^{2} - 1 ) ^{2}}{t ^{3}} d t

Expand

\frac{t ( 2 t ^{2} - 1 ) ^{2}}{t ^{3}} : 4 t^{\frac{3}{2}} - \frac{4}{t ^{\frac{1}{2}}} + \frac{1}{t ^{\frac{5}{2}}}

= \int 4 t^{\frac{3}{2}} - \frac{4}{t ^{\frac{1}{2}}} + \frac{1}{t ^{\frac{5}{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 4 t^{\frac{3}{2}} d t - \int \frac{4}{t ^{\frac{1}{2}}} d t + \int \frac{1}{t ^{\frac{5}{2}}} d t

\int 4 t^{\frac{3}{2}} d t = \frac{8}{5} t^{\frac{5}{2}}

\int \frac{4}{t ^{\frac{1}{2}}} d t = 8 t^{\frac{1}{2}}

\int \frac{1}{t ^{\frac{5}{2}}} d t = - \frac{2}{3 t ^{\frac{3}{2}}}

= \frac{8}{5} t^{\frac{5}{2}} - 8 t^{\frac{1}{2}} - \frac{2}{3 t ^{\frac{3}{2}}}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{8}{5} t^{\frac{5}{2}} - 8 t - \frac{2}{3 t ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{8}{5} t^{\frac{5}{2}} - 8 t - \frac{2}{3 t ^{\frac{3}{2}}} + C

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What is the integral of (sqrt(t)(2t^2-1)^2)/(t^3) ?
The integral of (sqrt(t)(2t^2-1)^2)/(t^3) is 8/5 t^{5/2}-8sqrt(t)-2/(3t^{3/2)}+C