What is the integral of (18t-5)/(sqrt(9t^2-5t+2)) ?

The integral of (18t-5)/(sqrt(9t^2-5t+2)) is 2sqrt(9t^2-5t+2)+C

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integral of (18t-5)/(sqrt(9t^2-5t+2))

\int \frac{18 t - 5}{9 t ^{2} - 5 t + 2} d t

2 9 t^{2} - 5 t + 2 + C

Solution steps

\int \frac{18 t - 5}{9 t ^{2} - 5 t + 2} d t

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 ((18 t - 5)^{2} + 47)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot 2 ((18 t - 5)^{2} + 47)^{\frac{1}{2}} : 2 9 t^{2} - 5 t + 2

= 2 9 t^{2} - 5 t + 2

Add a constant to the solution

= 2 9 t^{2} - 5 t + 2 + C

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What is the integral of (18t-5)/(sqrt(9t^2-5t+2)) ?
The integral of (18t-5)/(sqrt(9t^2-5t+2)) is 2sqrt(9t^2-5t+2)+C