What is the integral of 1/((x+5)sqrt(10x+x^2)) ?

The integral of 1/((x+5)sqrt(10x+x^2)) is 1/5 arcsec(1/5 x+1)+C

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integral of 1/((x+5)sqrt(10x+x^2))

\int \frac{1}{( x + 5 ) 10 x + x ^{2}} d x

\frac{1}{5} arcsec (\frac{1}{5} x + 1) + C

Solution steps

\int \frac{1}{( x + 5 ) 10 x + x ^{2}} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 25} d u

= \int \frac{1}{u u ^{2} - 25} d u

Apply Trigonometric Substitution

: \int \frac{1}{5} d v

= \int \frac{1}{5} d v

Take the constant out:

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

= \frac{1}{5} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot 1 \cdot v

Substitute back

= \frac{1}{5} \cdot 1 \cdot arcsec (\frac{1}{5} (x + 5))

Simplify

\frac{1}{5} \cdot 1 \cdot arcsec (\frac{1}{5} (x + 5)) : \frac{1}{5} arcsec (\frac{1}{5} x + 1)

= \frac{1}{5} arcsec (\frac{1}{5} x + 1)

Add a constant to the solution

= \frac{1}{5} arcsec (\frac{1}{5} x + 1) + C

\int 9 - x^{2} d x

- \frac{1}{x} 2

y = \frac{x}{x ^{2} + 1}

3 x^{2} - \frac{1}{x}

f (x) = x^{2} + 2 x, at x = 34

What is the integral of 1/((x+5)sqrt(10x+x^2)) ?
The integral of 1/((x+5)sqrt(10x+x^2)) is 1/5 arcsec(1/5 x+1)+C