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

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

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

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

- 2 arcsec (\frac{1}{2} (x + 3)) + x^{2} + 6 x + 5 + C

Solution steps

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

Apply u-substitution

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

Apply Trigonometric Substitution

= \int 2 tan^{2} (v) d v

Take the constant out:

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

= 2 \cdot \int tan^{2} (v) d v

Rewrite using trig identities

= 2 \cdot \int - 1 + sec^{2} (v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int 1 d v + \int sec^{2} (v) d v)

\int 1 d v = v

\int sec^{2} (v) d v = tan (v)

= 2 (- v + tan (v))

Substitute back

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

Simplify

2 (- arcsec (\frac{1}{2} (x + 3)) + tan (arcsec (\frac{1}{2} (x + 3)))) : - 2 arcsec (\frac{1}{2} (x + 3)) + x^{2} + 6 x + 5

= - 2 arcsec (\frac{1}{2} (x + 3)) + x^{2} + 6 x + 5

Add a constant to the solution

= - 2 arcsec (\frac{1}{2} (x + 3)) + x^{2} + 6 x + 5 + C

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