What is the integral of 1/(10+sqrt(x)) ?

The integral of 1/(10+sqrt(x)) is 2(10+sqrt(x)-10ln|10+sqrt(x)|)+C

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integral of 1/(10+sqrt(x))

\int \frac{1}{10 + x} d x

2 (10 + x - 10 ln 10 + x) + C

Solution steps

\int \frac{1}{10 + x} d x

Apply u-substitution

: \int \frac{2 ( u - 10 )}{u} d u

= \int \frac{2 ( u - 10 )}{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 - 10}{u} d u

Expand

\frac{u - 10}{u} : 1 - \frac{10}{u}

= 2 \cdot \int 1 - \frac{10}{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{10}{u} d u)

\int 1 d u = u

\int \frac{10}{u} d u = 10 ln ∣ u ∣

= 2 (u - 10 ln ∣ u ∣)

Substitute back

u = 10 + x

= 2 (10 + x - 10 ln 10 + x)

Add a constant to the solution

= 2 (10 + x - 10 ln 10 + x) + C

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What is the integral of 1/(10+sqrt(x)) ?
The integral of 1/(10+sqrt(x)) is 2(10+sqrt(x)-10ln|10+sqrt(x)|)+C