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

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

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

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

2 3 x^{2} - 5 x + 6 + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{3 u ^{2} + 47} d u

= \int \frac{u}{3 u ^{2} + 47} 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} + 47} 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} + 47)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot 2 ((6 x - 5)^{2} + 47)^{\frac{1}{2}} : 2 3 x^{2} - 5 x + 6

= 2 3 x^{2} - 5 x + 6

Add a constant to the solution

= 2 3 x^{2} - 5 x + 6 + C

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