Math / Calculus

Question $\int \frac{4 u-6}{u^{2}+6 u+18} d u$

Studdy Solution

STEP 1

1. The integral is of the form $\int \frac{a u + b}{u^2 + cu + d} \, du$ .
2. The numerator can potentially be expressed as a derivative of the denominator plus a constant.
3. The integral can be solved using substitution and partial fraction decomposition if necessary.

STEP 2

1. Simplify the integral by expressing the numerator in terms of the derivative of the denominator.
2. Perform substitution if necessary.
3. Integrate the resulting expression.

STEP 3

First, identify the derivative of the denominator $u^2 + 6u + 18$ . The derivative is:
$\frac{d}{du}(u^2 + 6u + 18) = 2u + 6$
Notice that the numerator $4u - 6$ can be expressed in terms of $2u + 6$ :
$4u - 6 = 2(2u + 6) - 18$
Thus, we can rewrite the integral as:
$\int \frac{4u - 6}{u^2 + 6u + 18} \, du = \int \frac{2(2u + 6) - 18}{u^2 + 6u + 18} \, du$

STEP 4

Split the integral into two separate integrals:
$\int \frac{2(2u + 6)}{u^2 + 6u + 18} \, du - \int \frac{18}{u^2 + 6u + 18} \, du$
The first integral can be simplified using substitution. Let $v = u^2 + 6u + 18$ , then $dv = (2u + 6) \, du$ .
For the first integral:
$\int \frac{2(2u + 6)}{u^2 + 6u + 18} \, du = \int \frac{2 \, dv}{v} = 2 \ln |v| + C_1 = 2 \ln |u^2 + 6u + 18| + C_1$

STEP 5

For the second integral, $\int \frac{18}{u^2 + 6u + 18} \, du$ , complete the square in the denominator:
$u^2 + 6u + 18 = (u + 3)^2 + 9$
This suggests a trigonometric substitution or recognizing it as a standard arctangent form:
$\int \frac{18}{(u + 3)^2 + 9} \, du = \int \frac{18}{(u + 3)^2 + 3^2} \, du$
Using the standard integral $\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C$ , we have:
$= 18 \cdot \frac{1}{3} \tan^{-1} \left(\frac{u + 3}{3}\right) + C_2 = 6 \tan^{-1} \left(\frac{u + 3}{3}\right) + C_2$

STEP 6

Combine the results from the two integrals:
$\int \frac{4u - 6}{u^2 + 6u + 18} \, du = 2 \ln |u^2 + 6u + 18| + 6 \tan^{-1} \left(\frac{u + 3}{3}\right) + C$
where $C = C_1 + C_2$ is the constant of integration.
The solution to the integral is:
$2 \ln |u^2 + 6u + 18| + 6 \tan^{-1} \left(\frac{u + 3}{3}\right) + C$

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