What is the integral of 1/(4sqrt(x)+x\sqrt{x)} ?

The integral of 1/(4sqrt(x)+x\sqrt{x)} is arctan(1/2 sqrt(x))+C

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integral of 1/(4sqrt(x)+x\sqrt{x)}

\int \frac{1}{4 x + x x} d x

arctan (\frac{1}{2} x) + C

Solution steps

\int \frac{1}{4 x + x x} d x

\frac{1}{4 x + x x} = \frac{4 x - x x}{( 4 x ) ^{2} - ( x x ) ^{2}}

= \int \frac{4 x - x x}{( 4 x ) ^{2} - ( x x ) ^{2}} d x

Simplify

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

= \int \frac{1}{x ( x + 4 )} d x

Apply u-substitution

: \int \frac{2}{u ^{2} + 4} d u

= \int \frac{2}{u ^{2} + 4} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

2 \cdot \frac{1}{2} arctan (\frac{x}{2}) : arctan (\frac{1}{2} x)

= arctan (\frac{1}{2} x)

Add a constant to the solution

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

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What is the integral of 1/(4sqrt(x)+x\sqrt{x)} ?
The integral of 1/(4sqrt(x)+x\sqrt{x)} is arctan(1/2 sqrt(x))+C