What is the integral of 3/(sqrt(x)+x\sqrt{x)} ?

The integral of 3/(sqrt(x)+x\sqrt{x)} is 6arctan(sqrt(x))+C

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integral of 3/(sqrt(x)+x\sqrt{x)}

\int \frac{3}{x + x x} d x

6 arctan (x) + C

Solution steps

\int \frac{3}{x + x x} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x + x x} d x

Apply u-substitution

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

= 3 \cdot \int \frac{2}{1 + u ^{2}} d u

Take the constant out:

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

= 3 \cdot 2 \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 3 \cdot 2 arctan (u)

Substitute back

u = x

= 3 \cdot 2 arctan (x)

Simplify

= 6 arctan (x)

Add a constant to the solution

= 6 arctan (x) + C

\int \frac{2 x ^{2} - x + 4}{x ^{3} + 4 x} d x

\int x in (x) d x

\int \frac{cos ( x )}{1 - sin ( x )} d x

(e^{3 x})^{^{'}}

x \to 3 lim (\frac{2 x - 3}{x + 5})

What is the integral of 3/(sqrt(x)+x\sqrt{x)} ?
The integral of 3/(sqrt(x)+x\sqrt{x)} is 6arctan(sqrt(x))+C