What is the integral of (x^{0.5})/((1+x)^2) ?

The integral of (x^{0.5})/((1+x)^2) is -(sqrt(x))/(1+x)+arctan(sqrt(x))+C

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integral of (x^{0.5})/((1+x)^2)

integral

\frac{x ^{0.5}}{( 1 + x ) ^{2}}

- \frac{x}{1 + x} + arctan (x) + C

Solution steps

\int \frac{x ^{0.5}}{( 1 + x ) ^{2}} d x

Simplify

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

= \int \frac{x ^{0.5}}{( 1 + x ) ( 1 + x )} d x

(1 + x) (1 + x) = (1 + x)^{2}

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= - \frac{x}{1 + x} + arctan (x)

Add a constant to the solution

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

\frac{x}{( ( x - 1 ) ( x - 2 ) ( x - 3 ) )}

\frac{cot ( x )}{2}

(a - x^{2})^{\frac{1}{2}}

sin^{3} (x) cos^{3} (x)

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

What is the integral of (x^{0.5})/((1+x)^2) ?
The integral of (x^{0.5})/((1+x)^2) is -(sqrt(x))/(1+x)+arctan(sqrt(x))+C