Math / Calculus

Question $\int \frac{2 x+5}{3 x-1} d x$

Studdy Solution

STEP 1

1. The given integral is $\int \frac{2x + 5}{3x - 1} \, dx$ .
2. We will use a method of substitution for solving this integral.
3. We assume the integrand can be simplified by recognizing it as a rational function that may benefit from a substitution.
4. The substitution will simplify the integral into a form that can be easily integrated.

STEP 2

1. Make an appropriate substitution to simplify the integrand.
2. Simplify the integral after substitution.
3. Integrate the simplified function.
4. Substitute back to the original variable.

STEP 3

Make the substitution $u = 3x - 1$ . Then, compute the differential $du$ .
$u = 3x - 1 \implies du = 3 \, dx \implies dx = \frac{du}{3}$

STEP 4

Rewrite the integral $\int \frac{2x + 5}{3x - 1} \, dx$ in terms of $u$ .
$\int \frac{2x + 5}{3x - 1} \, dx = \int \frac{2x + 5}{u} \cdot \frac{du}{3}$

STEP 5

Express $x$ in terms of $u$ to simplify the integrand further. From the substitution $u = 3x - 1$ , we get $x = \frac{u + 1}{3}$ .
$2x + 5 = 2\left(\frac{u + 1}{3}\right) + 5 = \frac{2u + 2}{3} + 5 = \frac{2u + 2 + 15}{3} = \frac{2u + 17}{3}$

STEP 6

Substitute this back into the integral.
$\int \frac{2x + 5}{u} \cdot \frac{du}{3} = \int \frac{\frac{2u + 17}{3}}{u} \cdot \frac{du}{3} = \int \frac{2u + 17}{3u} \cdot \frac{du}{3}$

STEP 7

Simplify the integrand.
$\int \frac{2u + 17}{3u} \cdot \frac{du}{3} = \int \frac{2u + 17}{3u \cdot 3} \, du = \int \frac{2u + 17}{9u} \, du = \int \left( \frac{2u}{9u} + \frac{17}{9u} \right) \, du = \int \left( \frac{2}{9} + \frac{17}{9u} \right) \, du$

STEP 8

Integrate the simplified function.
$\int \left( \frac{2}{9} + \frac{17}{9u} \right) \, du = \int \frac{2}{9} \, du + \int \frac{17}{9u} \, du = \frac{2}{9} \int \, du + \frac{17}{9} \int \frac{1}{u} \, du$

STEP 9

Compute the integrals.
$\frac{2}{9} \int \, du + \frac{17}{9} \int \frac{1}{u} \, du = \frac{2}{9} u + \frac{17}{9} \ln|u| + C$

STEP 10

Substitute back $u = 3x - 1$ .
$\frac{2}{9} (3x - 1) + \frac{17}{9} \ln|3x - 1| + C$

STEP 11

Simplify the final expression.
$\frac{2}{9} (3x - 1) + \frac{17}{9} \ln|3x - 1| + C = \frac{2}{3} x - \frac{2}{9} + \frac{17}{9} \ln|3x - 1| + C$
The solution to the integral is:
$\int \frac{2x + 5}{3x - 1} \, dx = \frac{2}{3} x - \frac{2}{9} + \frac{17}{9} \ln|3x - 1| + C$

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