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

The integral of ((1+x))/(1+sqrt(x)) is 2/3 x^{3/2}-x+4sqrt(x)-4ln|sqrt(x)+1|+C

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

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

\frac{2}{3} x^{\frac{3}{2}} - x + 4 x - 4 ln x + 1 + C

Solution steps

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

Apply u-substitution

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

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

Long division

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

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

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

= 2 (\frac{u ^{3}}{3} - \frac{u ^{2}}{2} + 2 u - 2 ln ∣ u + 1 ∣)

Substitute back

u = x

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

Simplify

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

= \frac{2}{3} x^{\frac{3}{2}} - x + 4 x - 4 ln x + 1

Add a constant to the solution

= \frac{2}{3} x^{\frac{3}{2}} - x + 4 x - 4 ln x + 1 + C

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What is the integral of ((1+x))/(1+sqrt(x)) ?
The integral of ((1+x))/(1+sqrt(x)) is 2/3 x^{3/2}-x+4sqrt(x)-4ln|sqrt(x)+1|+C