Math / Calculus

QuestionEvaluate $\int_{0}^{1} \frac{y \ln ^{4}\left(y^{2}+1\right)}{y^{2}+1} d y$

Studdy Solution

STEP 1

1. The integral is definite with limits from 0 to 1.
2. The integrand involves a logarithmic function and a rational function.
3. We may need to use substitution to simplify the integral.

STEP 2

1. Identify a suitable substitution to simplify the integral.
2. Change the limits of integration according to the substitution.
3. Evaluate the new integral.
4. Substitute back if necessary and compute the definite integral.

STEP 3

Identify a suitable substitution. Notice that the integrand contains $\ln^4(y^2 + 1)$ and $y^2 + 1$ in the denominator. A good substitution is $u = y^2 + 1$ , which simplifies the expression:
Let $u = y^2 + 1$ . Then, $du = 2y \, dy$ , or $y \, dy = \frac{1}{2} du$ .

STEP 4

Change the limits of integration according to the substitution:
When $y = 0$ , $u = 0^2 + 1 = 1$ .
When $y = 1$ , $u = 1^2 + 1 = 2$ .
The new limits of integration are from $u = 1$ to $u = 2$ .

STEP 5

Substitute into the integral:
The integral becomes:
$\int_{1}^{2} \frac{\ln^4(u)}{u} \cdot \frac{1}{2} \, du$
Simplify the integral:
$\frac{1}{2} \int_{1}^{2} \frac{\ln^4(u)}{u} \, du$

STEP 6

Evaluate the integral:
The integral $\int \frac{\ln^4(u)}{u} \, du$ can be evaluated using the substitution $v = \ln(u)$ , which gives $dv = \frac{1}{u} \, du$ . Thus, the integral becomes:
$\int v^4 \, dv$
Evaluate this integral:
$\int v^4 \, dv = \frac{v^5}{5} + C$
Substitute back $v = \ln(u)$ :
$\frac{\ln^5(u)}{5}$
Now evaluate the definite integral from 1 to 2:
$\frac{1}{2} \left[ \frac{\ln^5(u)}{5} \right]_{1}^{2}$

STEP 7

Compute the definite integral:
$\frac{1}{2} \left( \frac{\ln^5(2)}{5} - \frac{\ln^5(1)}{5} \right)$
Since $\ln(1) = 0$ , the expression simplifies to:
$\frac{1}{2} \cdot \frac{\ln^5(2)}{5} = \frac{\ln^5(2)}{10}$
The value of the integral is:
$\boxed{\frac{\ln^5(2)}{10}}$

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