What is the integral of (15-18t)/(3t^2-5t+13) ?

The integral of (15-18t)/(3t^2-5t+13) is -3ln|36t^2-60t+156|+C

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integral of (15-18t)/(3t^2-5t+13)

\int \frac{15 - 18 t}{3 t ^{2} - 5 t + 13} d t

- 3 ln 36 t^{2} - 60 t + 156 + C

Solution steps

\int \frac{15 - 18 t}{3 t ^{2} - 5 t + 13} d t

Complete the square

3 t^{2} - 5 t + 13 : 3 (t - \frac{5}{6})^{2} + \frac{131}{12}

= \int \frac{15 - 18 t}{3 ( t - \frac{5}{6} ) ^{2} + \frac{131}{12}} d t

Apply u-substitution

: \int - \frac{216 u}{36 u ^{2} + 131} d u

= \int - \frac{216 u}{36 u ^{2} + 131} d u

Take the constant out:

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

= - 216 \cdot \int \frac{u}{36 u ^{2} + 131} d u

Apply u-substitution

: \int \frac{1}{72 v} d v

= - 216 \cdot \int \frac{1}{72 v} d v

Take the constant out:

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

= - 216 \cdot \frac{1}{72} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - 216 \cdot \frac{1}{72} ln ∣ v ∣

Substitute back

= - 216 \cdot \frac{1}{72} ln 36 (t - \frac{5}{6})^{2} + 131

Simplify

- 216 \cdot \frac{1}{72} ln 36 (t - \frac{5}{6})^{2} + 131 : - 3 ln 36 t^{2} - 60 t + 156

= - 3 ln 36 t^{2} - 60 t + 156

Add a constant to the solution

= - 3 ln 36 t^{2} - 60 t + 156 + C

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What is the integral of (15-18t)/(3t^2-5t+13) ?
The integral of (15-18t)/(3t^2-5t+13) is -3ln|36t^2-60t+156|+C