What is the integral of xsqrt((1-x)/(1+x)) ?

The integral of xsqrt((1-x)/(1+x)) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xsqrt((1-x)/(1+x))

\int x \frac{1 - x}{1 + x} d x

arctan (\frac{- x + 1}{x + 1}) + \frac{1}{8} sin (4 arctan (\frac{- x + 1}{x + 1})) - \frac{5}{4} - x + 1 x + 1 + \frac{1}{4} - x + 1 (x + 1)^{\frac{3}{2}} + C

Solution steps

\int x \frac{1 - x}{1 + x} d x

Apply u-substitution

: \int \frac{4 u ^{2} ( u ^{2} - 1 )}{( u ^{2} + 1 ) ^{\frac{5}{2}} u ^{2} + 1} d u

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

Take the constant out:

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

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

(u^{2} + 1)^{\frac{5}{2}} = (u^{2} + 1)^{2} u^{2} + 1

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

(u^{2} + 1)^{2} u^{2} + 1 u^{2} + 1 = (u^{2} + 1)^{3}

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

Expand

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

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

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

\int \frac{u ^{4}}{( u ^{2} + 1 ) ^{3}} d u = \frac{3}{8} arctan (u) - \frac{5}{8} \cdot \frac{u}{1 + u ^{2}} + \frac{u}{4 ( u ^{2} + 1 ) ^{2}}

\int \frac{u ^{2}}{( u ^{2} + 1 ) ^{3}} d u = \frac{1}{8} (arctan (u) - \frac{1}{4} sin (4 arctan (u)))

= 4 (\frac{3}{8} arctan (u) - \frac{5}{8} \cdot \frac{u}{1 + u ^{2}} + \frac{u}{4 ( u ^{2} + 1 ) ^{2}} - \frac{1}{8} (arctan (u) - \frac{1}{4} sin (4 arctan (u))))

Substitute back

u = \frac{1 - x}{1 + x}

= 4 \frac{3}{8} arctan (\frac{1 - x}{1 + x}) - \frac{5}{8} \cdot \frac{\frac{1 - x}{1 + x}}{1 + ( \frac{1 - x}{1 + x} ) ^{2}} + \frac{\frac{1 - x}{1 + x}}{4 ( ( \frac{1 - x}{1 + x} ) ^{2} + 1 ) ^{2}} - \frac{1}{8} (arctan (\frac{1 - x}{1 + x}) - \frac{1}{4} sin (4 arctan (\frac{1 - x}{1 + x})))

Simplify

4 \frac{3}{8} arctan (\frac{1 - x}{1 + x}) - \frac{5}{8} \cdot \frac{\frac{1 - x}{1 + x}}{1 + ( \frac{1 - x}{1 + x} ) ^{2}} + \frac{\frac{1 - x}{1 + x}}{4 ( ( \frac{1 - x}{1 + x} ) ^{2} + 1 ) ^{2}} - \frac{1}{8} (arctan (\frac{1 - x}{1 + x}) - \frac{1}{4} sin (4 arctan (\frac{1 - x}{1 + x}))) : arctan (\frac{- x + 1}{x + 1}) + \frac{1}{8} sin (4 arctan (\frac{- x + 1}{x + 1})) - \frac{5}{4} - x + 1 x + 1 + \frac{1}{4} - x + 1 (x + 1)^{\frac{3}{2}}

= arctan (\frac{- x + 1}{x + 1}) + \frac{1}{8} sin (4 arctan (\frac{- x + 1}{x + 1})) - \frac{5}{4} - x + 1 x + 1 + \frac{1}{4} - x + 1 (x + 1)^{\frac{3}{2}}

Add a constant to the solution

= arctan (\frac{- x + 1}{x + 1}) + \frac{1}{8} sin (4 arctan (\frac{- x + 1}{x + 1})) - \frac{5}{4} - x + 1 x + 1 + \frac{1}{4} - x + 1 (x + 1)^{\frac{3}{2}} + C

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

\int sec^{11} (x) tan (x) d x

\int - 3 x^{3} d x

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

\int 4 x + 7 d x

What is the integral of xsqrt((1-x)/(1+x)) ?
The integral of xsqrt((1-x)/(1+x)) is