What is the integral of (t^3)/((4t^2+9)^{3/2)} ?

The integral of (t^3)/((4t^2+9)^{3/2)} is (2t^2+9)/(8(4t^2+9)^{1/2)}+C

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integral of (t^3)/((4t^2+9)^{3/2)}

\int \frac{t ^{3}}{( 4 t ^{2} + 9 ) ^{\frac{3}{2}}} d t

\frac{2 t ^{2} + 9}{8 ( 4 t ^{2} + 9 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{t ^{3}}{( 4 t ^{2} + 9 ) ^{\frac{3}{2}}} d t

Apply u-substitution

= \int \frac{u - 9}{32 u ^{\frac{3}{2}}} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int \frac{u - 9}{u ^{\frac{3}{2}}} d u

Expand

\frac{u - 9}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} - \frac{9}{u ^{\frac{3}{2}}}

= \frac{1}{32} \cdot \int \frac{1}{u ^{\frac{1}{2}}} - \frac{9}{u ^{\frac{3}{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}{32} (\int \frac{1}{u ^{\frac{1}{2}}} d u - \int \frac{9}{u ^{\frac{3}{2}}} d u)

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

\int \frac{9}{u ^{\frac{3}{2}}} d u = - \frac{18}{u ^{\frac{1}{2}}}

= \frac{1}{32} (2 u^{\frac{1}{2}} - (- \frac{18}{u ^{\frac{1}{2}}}))

Substitute back

u = 4 t^{2} + 9

= \frac{1}{32} (2 (4 t^{2} + 9)^{\frac{1}{2}} - (- \frac{18}{( 4 t ^{2} + 9 ) ^{\frac{1}{2}}}))

Simplify

\frac{1}{32} (2 (4 t^{2} + 9)^{\frac{1}{2}} - (- \frac{18}{( 4 t ^{2} + 9 ) ^{\frac{1}{2}}})) : \frac{2 t ^{2} + 9}{8 ( 4 t ^{2} + 9 ) ^{\frac{1}{2}}}

= \frac{2 t ^{2} + 9}{8 ( 4 t ^{2} + 9 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{2 t ^{2} + 9}{8 ( 4 t ^{2} + 9 ) ^{\frac{1}{2}}} + C

What is the integral of (t^3)/((4t^2+9)^{3/2)} ?
The integral of (t^3)/((4t^2+9)^{3/2)} is (2t^2+9)/(8(4t^2+9)^{1/2)}+C