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

The integral of (x+3)/((x^2+6x)^{1/2)} is (x^2+6x)^{1/2}+C

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integral of (x+3)/((x^2+6x)^{1/2)}

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Power Rule

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} + 6 x

= \frac{1}{2} \cdot 2 (x^{2} + 6 x)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (x^{2} + 6 x)^{\frac{1}{2}} : (x^{2} + 6 x)^{\frac{1}{2}}

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

Add a constant to the solution

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

What is the integral of (x+3)/((x^2+6x)^{1/2)} ?
The integral of (x+3)/((x^2+6x)^{1/2)} is (x^2+6x)^{1/2}+C