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

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

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

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

2 x (3 x + 5) + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{3 u ^{2} - 25} d u

= \int \frac{u}{3 u ^{2} - 25} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{u}{u ^{2} - 25} d u

Apply u-substitution

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

= \frac{1}{3} \cdot \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= \frac{1}{3} \cdot \frac{1}{2} \cdot 2 v^{\frac{1}{2}}

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} \cdot 2 ((6 x + 5)^{2} - 25)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot 2 ((6 x + 5)^{2} - 25)^{\frac{1}{2}} : 2 x (3 x + 5)

= 2 x (3 x + 5)

Add a constant to the solution

= 2 x (3 x + 5) + C

\int_{3}^{5} \frac{1}{x ^{2} - 9} d x

f (x) = (x^{3} - 10 x) (3 x^{2} + 4 x)

y = x^{2} ln (x), y = 4 ln (x)

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

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

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