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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = 4 - t^{2}

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

Simplify

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

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

Add a constant to the solution

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

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