What is the integral of 1/((4x-x^2)^{3/2)} ?

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

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integral of 1/((4x-x^2)^{3/2)}

\int \frac{1}{( 4 x - x ^{2} ) ^{\frac{3}{2}}} d x

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

Solution steps

\int \frac{1}{( 4 x - x ^{2} ) ^{\frac{3}{2}}} d x

Complete the square

4 x - x^{2} : - (x - 2)^{2} + 4

= \int \frac{1}{( - ( x - 2 ) ^{2} + 4 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u

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

Apply Trigonometric Substitution

: \int \frac{1}{4 cos ^{2} ( v )} d v

= \int \frac{1}{4 cos ^{2} ( v )} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{cos ^{2} ( v )} d v

Use the common integral:

\int \frac{1}{cos ^{2} ( v )} d v = tan (v)

= \frac{1}{4} tan (v)

Substitute back

= \frac{1}{4} tan (arcsin (\frac{1}{2} (x - 2)))

Simplify

\frac{1}{4} tan (arcsin (\frac{1}{2} (x - 2))) : \frac{x - 2}{4 ( - x ^{2} + 4 x ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

n \to 0 - lim (5 + \frac{4}{n})

t e^{- 3 t} sin (5 t)

\int \frac{2 x - 1}{x ^{2} + 4} d x

\int \frac{4}{5} x^{3} d x

i (t)

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