What is the integral of (32)/(sqrt(x)+x\sqrt{x)} ?

The integral of (32)/(sqrt(x)+x\sqrt{x)} is 64arctan(sqrt(x))+C

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integral of (32)/(sqrt(x)+x\sqrt{x)}

\int \frac{32}{x + x x} d x

64 arctan (x) + C

Solution steps

\int \frac{32}{x + x x} d x

Take the constant out:

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

= 32 \cdot \int \frac{1}{x + x x} d x

Apply u-substitution

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

= 32 \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

= 32 \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)

= 32 \cdot 2 arctan (u)

Substitute back

u = x

= 32 \cdot 2 arctan (x)

Simplify

= 64 arctan (x)

Add a constant to the solution

= 64 arctan (x) + C

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What is the integral of (32)/(sqrt(x)+x\sqrt{x)} ?
The integral of (32)/(sqrt(x)+x\sqrt{x)} is 64arctan(sqrt(x))+C