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

The integral of 1/((1+sqrt(x))) is 2(1+sqrt(x)-ln|1+sqrt(x)|)+C

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integral of 1/((1+sqrt(x)))

\int \frac{1}{( 1 + x )} d x

2 (1 + x - ln 1 + x) + C

Solution steps

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 2 \cdot \int 1 - \frac{1}{u} 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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 (u - ln ∣ u ∣)

Substitute back

u = 1 + x

= 2 (1 + x - ln 1 + x)

Add a constant to the solution

= 2 (1 + x - ln 1 + x) + C

\int_{0}^{π} sin (2 x) cos (x) d x

y^{^{''}} - y^{^{'}} = 3 x^{2} e^{x}

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

\int \frac{1}{( x - 9 ) x ^{2} - 18 x} d x

\frac{2}{( - x ^{3} + 9 x ) ^{4}}

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