What is the integral of 1/((x+3)sqrt(x^2+6x+5)) ?

The integral of 1/((x+3)sqrt(x^2+6x+5)) is 1/2 arcsec(1/2 (x+3))+C

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integral of 1/((x+3)sqrt(x^2+6x+5))

\int \frac{1}{( x + 3 ) x ^{2} + 6 x + 5} d x

\frac{1}{2} arcsec (\frac{1}{2} (x + 3)) + C

Solution steps

\int \frac{1}{( x + 3 ) x ^{2} + 6 x + 5} d x

Apply u-substitution

= \int \frac{1}{u u ^{2} - 4} d u

Apply Trigonometric Substitution

= \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (x + 3))

Simplify

= \frac{1}{2} arcsec (\frac{1}{2} (x + 3))

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{1}{2} (x + 3)) + C

What is the integral of 1/((x+3)sqrt(x^2+6x+5)) ?
The integral of 1/((x+3)sqrt(x^2+6x+5)) is 1/2 arcsec(1/2 (x+3))+C