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

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

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

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

- \frac{t + 1}{( 2 t + 3 ) ^{\frac{3}{2}}} + C

Solution steps

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

Apply u-substitution

= \int \frac{u - 3}{4 u ^{\frac{5}{2}}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u - 3}{u ^{\frac{5}{2}}} d u

Expand

\frac{u - 3}{u ^{\frac{5}{2}}} : \frac{1}{u ^{\frac{3}{2}}} - \frac{3}{u ^{\frac{5}{2}}}

= \frac{1}{4} \cdot \int \frac{1}{u ^{\frac{3}{2}}} - \frac{3}{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}{4} (\int \frac{1}{u ^{\frac{3}{2}}} d u - \int \frac{3}{u ^{\frac{5}{2}}} d u)

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

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

= \frac{1}{4} (- \frac{2}{u ^{\frac{1}{2}}} - (- \frac{2}{u ^{\frac{3}{2}}}))

Substitute back

u = 2 t + 3

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

Simplify

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

= - \frac{t + 1}{( 2 t + 3 ) ^{\frac{3}{2}}}

Add a constant to the solution

= - \frac{t + 1}{( 2 t + 3 ) ^{\frac{3}{2}}} + C

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