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

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

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

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

2 arctan (x) + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = x

= 2 arctan (x)

Add a constant to the solution

= 2 arctan (x) + C

\int_{1}^{x} \frac{1}{t} d t

x^{6 x}

\frac{x ^{3}}{3} + x^{2} + 4 x - 10

\frac{d}{d x} (\frac{3 x ^{2}}{( x ^{2} + x ) ^{3}})

17 e^{0.02 t}, 21 e^{0.02 t}, [0, 50]

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