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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Complete the square

2 x - x^{2} : - (x - 1)^{2} + 1

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec^{2} (v) d v

= \int sec^{2} (v) d v

Use the common integral:

\int sec^{2} (v) d v = tan (v)

= tan (v)

Substitute back

= tan (arcsin (x - 1))

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( 9 x ^{2} + 1 ) ^{2}} d x

\int (sin (6 x) + cos (\frac{1}{3}) x) d x

\int e^{- \frac{2}{3} x} d x

\int 4 x^{3} - x d x

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