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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 u}{1 + u} d u

= \int \frac{2 u}{1 + u} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{1 + u} d u

Apply u-substitution

: \int \frac{v - 1}{v} d v

= 2 \cdot \int \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 2 \cdot \int 1 - \frac{1}{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int 1 d v - \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (v - ln ∣ v ∣)

Substitute back

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

Add a constant to the solution

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

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

\frac{1}{( 2 + sin ( 3 x ) )}

\frac{1}{e ^{x} ( 1 + e ^{x} )}

x^{3} sin (x^{2})

e^{\frac{x}{3}}

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