Studdy AI Logo
Math  /  Calculus

Question∫1v(Ln⁡v−1)dv\int \frac{1}{v(\operatorname{Ln} v-1)} d v

Studdy Solution

STEP 1

1. We are given the integral ∫1v(ln⁡v−1) dv\int \frac{1}{v(\ln v - 1)} \, dv.
2. We need to find the antiderivative of the given function.

STEP 2

1. Simplify the integrand if possible.
2. Use substitution to simplify the integral.
3. Solve the integral using the substitution.
4. Back-substitute to express the solution in terms of the original variable.

STEP 3

Examine the integrand 1v(ln⁡v−1)\frac{1}{v(\ln v - 1)} to see if it can be simplified directly. In this case, direct simplification is not possible, so we proceed to substitution.

STEP 4

Use substitution to simplify the integral. Let u=ln⁡v−1 u = \ln v - 1 . Then, the derivative du=1v dv du = \frac{1}{v} \, dv .

STEP 5

Rewrite the integral in terms of u u :
∫1v(ln⁡v−1) dv=∫1u⋅v⋅1v dv=∫1u⋅du\int \frac{1}{v(\ln v - 1)} \, dv = \int \frac{1}{u} \cdot v \cdot \frac{1}{v} \, dv = \int \frac{1}{u} \cdot du
This simplifies to:
∫1u du\int \frac{1}{u} \, du

STEP 6

Solve the integral ∫1u du\int \frac{1}{u} \, du:
∫1u du=ln⁡∣u∣+C\int \frac{1}{u} \, du = \ln |u| + C
where C C is the constant of integration.

STEP 7

Back-substitute u=ln⁡v−1 u = \ln v - 1 into the solution:
ln⁡∣u∣+C=ln⁡∣ln⁡v−1∣+C\ln |u| + C = \ln |\ln v - 1| + C
The solution to the integral is:
ln⁡∣ln⁡v−1∣+C\boxed{\ln |\ln v - 1| + C}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord