What is the integral of 1/(sqrt(x)(49+x)) ?

The integral of 1/(sqrt(x)(49+x)) is 2/7 arctan(1/7 sqrt(x))+C

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integral of 1/(sqrt(x)(49+x))

\int \frac{1}{x ( 49 + x )} d x

\frac{2}{7} arctan (\frac{1}{7} x) + C

Solution steps

\int \frac{1}{x ( 49 + x )} d x

Apply u-substitution

= \int \frac{2}{49 + 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}{49 + u ^{2}} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{7 ( 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}{7} \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}{7} arctan (v)

Substitute back

= 2 \cdot \frac{1}{7} arctan (\frac{x}{7})

Simplify

2 \cdot \frac{1}{7} arctan (\frac{x}{7}) : \frac{2}{7} arctan (\frac{1}{7} x)

= \frac{2}{7} arctan (\frac{1}{7} x)

Add a constant to the solution

= \frac{2}{7} arctan (\frac{1}{7} x) + C

What is the integral of 1/(sqrt(x)(49+x)) ?
The integral of 1/(sqrt(x)(49+x)) is 2/7 arctan(1/7 sqrt(x))+C