What is the integral of (x+5)/(sqrt(3x^2+30x-4)) ?

The integral of (x+5)/(sqrt(3x^2+30x-4)) is 1/3 sqrt(3x^2+30x-4)+C

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integral of (x+5)/(sqrt(3x^2+30x-4))

\int \frac{x + 5}{3 x ^{2} + 30 x - 4} d x

\frac{1}{3} 3 x^{2} + 30 x - 4 + C

Solution steps

\int \frac{x + 5}{3 x ^{2} + 30 x - 4} d x

Apply u-substitution

= \int \frac{u}{3 u ^{2} - 79} d u

Apply u-substitution

= \int \frac{1}{6 v} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{v} d v

Apply the Power Rule

= \frac{1}{6} \cdot 2 v^{\frac{1}{2}}

Substitute back

= \frac{1}{6} \cdot 2 (3 (x + 5)^{2} - 79)^{\frac{1}{2}}

Simplify

\frac{1}{6} \cdot 2 (3 (x + 5)^{2} - 79)^{\frac{1}{2}} : \frac{1}{3} 3 x^{2} + 30 x - 4

= \frac{1}{3} 3 x^{2} + 30 x - 4

Add a constant to the solution

= \frac{1}{3} 3 x^{2} + 30 x - 4 + C

What is the integral of (x+5)/(sqrt(3x^2+30x-4)) ?
The integral of (x+5)/(sqrt(3x^2+30x-4)) is 1/3 sqrt(3x^2+30x-4)+C