What is the integral of t^3(25+81t^2)^{5/2} ?

The integral of t^3(25+81t^2)^{5/2} is 1/413343 (567t^2-50)(81t^2+25)^{7/2}+C

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integral of t^3(25+81t^2)^{5/2}

\int t^{3} (25 + 81 t^{2})^{\frac{5}{2}} d t

\frac{1}{413343} (567 t^{2} - 50) (81 t^{2} + 25)^{\frac{7}{2}} + C

Solution steps

\int t^{3} (25 + 81 t^{2})^{\frac{5}{2}} d t

Apply u-substitution

: \int \frac{u ^{\frac{5}{2}} ( u - 25 )}{13122} d u

= \int \frac{u ^{\frac{5}{2}} ( u - 25 )}{13122} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{13122} \cdot \int u^{\frac{5}{2}} (u - 25) d u

Expand

u^{\frac{5}{2}} (u - 25) : u^{\frac{7}{2}} - 25 u^{\frac{5}{2}}

= \frac{1}{13122} \cdot \int u^{\frac{7}{2}} - 25 u^{\frac{5}{2}} d u

Apply the Sum Rule:

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

= \frac{1}{13122} (\int u^{\frac{7}{2}} d u - \int 25 u^{\frac{5}{2}} d u)

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

\int 25 u^{\frac{5}{2}} d u = \frac{50}{7} u^{\frac{7}{2}}

= \frac{1}{13122} (\frac{2}{9} u^{\frac{9}{2}} - \frac{50}{7} u^{\frac{7}{2}})

Substitute back

u = 25 + 81 t^{2}

= \frac{1}{13122} (\frac{2}{9} (25 + 81 t^{2})^{\frac{9}{2}} - \frac{50}{7} (25 + 81 t^{2})^{\frac{7}{2}})

Simplify

\frac{1}{13122} (\frac{2}{9} (25 + 81 t^{2})^{\frac{9}{2}} - \frac{50}{7} (25 + 81 t^{2})^{\frac{7}{2}}) : \frac{1}{413343} (567 t^{2} - 50) (81 t^{2} + 25)^{\frac{7}{2}}

= \frac{1}{413343} (567 t^{2} - 50) (81 t^{2} + 25)^{\frac{7}{2}}

Add a constant to the solution

= \frac{1}{413343} (567 t^{2} - 50) (81 t^{2} + 25)^{\frac{7}{2}} + C

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What is the integral of t^3(25+81t^2)^{5/2} ?
The integral of t^3(25+81t^2)^{5/2} is 1/413343 (567t^2-50)(81t^2+25)^{7/2}+C