Math / Calculus

Question $\int \frac{\ln ^{2} x+2 \ln x-1}{x} d x$

Studdy Solution

STEP 1

1. The integral is of the form $\int \frac{f(x)}{x} \, dx$ , which suggests a potential substitution method.
2. The function in the numerator, $\ln^2 x + 2 \ln x - 1$ , can be simplified using substitution.

STEP 2

1. Choose an appropriate substitution.
2. Simplify the integral using the substitution.
3. Integrate the simplified expression.
4. Substitute back to the original variable.

STEP 3

Choose a substitution that simplifies the integral. Let $u = \ln x$ . Then, $du = \frac{1}{x} dx$ .

STEP 4

Substitute $u = \ln x$ and $du = \frac{1}{x} dx$ into the integral:
$\int \frac{\ln^2 x + 2 \ln x - 1}{x} \, dx = \int (u^2 + 2u - 1) \, du$

STEP 5

Integrate the polynomial expression with respect to $u$ :
$\int (u^2 + 2u - 1) \, du = \int u^2 \, du + \int 2u \, du - \int 1 \, du$
Calculate each integral separately:
$\int u^2 \, du = \frac{u^3}{3}$ $\int 2u \, du = u^2$ $\int 1 \, du = u$
Combine the results:
$\frac{u^3}{3} + u^2 - u + C$
where $C$ is the constant of integration.

STEP 6

Substitute back $u = \ln x$ into the integrated expression:
$\frac{(\ln x)^3}{3} + (\ln x)^2 - \ln x + C$
The solution to the integral is:
$\boxed{\frac{(\ln x)^3}{3} + (\ln x)^2 - \ln x + C}$

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