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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} - 9}{u} d u

= \int \frac{u ^{2} - 9}{u} d u

Apply Trigonometric Substitution

: \int 3 tan^{2} (v) d v

= \int 3 tan^{2} (v) d v

Take the constant out:

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

= 3 \cdot \int tan^{2} (v) d v

Rewrite using trig identities

= 3 \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

= 3 (- \int 1 d v + \int sec^{2} (v) d v)

\int 1 d v = v

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

= 3 (- v + tan (v))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int (0.1 x + 1) e^{0.03 x} d x

f (x) = (x - 6) (3 x + 5)

\frac{d}{d x} (e^{2 x})

\frac{d}{d x} ((x^{3} - 3 x^{2} + 1)^{- 3})

x^{3} - 9 x^{2} + 10 x, at = - 2

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