Math  /  Calculus

QuestionEvaluate the integral. 3x3/2xdx3x3/2xdx=\frac{\int \frac{3}{x^{3 / 2}-\sqrt{x}} d x}{\int \frac{3}{x^{3 / 2}-\sqrt{x}} d x=\square} \square

Studdy Solution

STEP 1

What is this asking? We need to find the antiderivative of a fraction involving square roots and fractional exponents of xx. Watch out! Don't forget to add the constant of integration, it's super important!
Also, fractional exponents can be tricky, so let's be extra careful with them.

STEP 2

1. Simplify the integrand
2. Use substitution
3. Integrate
4. Substitute back
5. Simplify

STEP 3

Alright, let's **rewrite** that integrand to make it easier to work with!
We can express both terms in the denominator with fractional exponents.
Remember, x\sqrt{x} is the same as x12x^{\frac{1}{2}}, and we've already got x32x^{\frac{3}{2}} down there.
So, our integral becomes: 3x32x12dx\int \frac{3}{x^{\frac{3}{2}} - x^{\frac{1}{2}}} \, dx

STEP 4

Now, let's **factor out** x12x^{\frac{1}{2}} from the denominator.
This will help us simplify things even further! 3x12(x1)dx\int \frac{3}{x^{\frac{1}{2}}(x - 1)} \, dx

STEP 5

We can **rewrite** the integrand one more time to make it even clearer: 3x12x1dx\int \frac{3x^{-\frac{1}{2}}}{x - 1} \, dx Look at that, much better!

STEP 6

Let's **use substitution** to make this integral easier to handle.
We'll let u=x1u = x - 1.
This means that x=u+1x = u + 1, and du=dxdu = dx.
Don't forget the chain rule!
Now, we can **rewrite** our integral in terms of uu: 3(u+1)12udu\int \frac{3(u + 1)^{-\frac{1}{2}}}{u} \, du

STEP 7

Unfortunately, this integral doesn't have a nice, clean solution in terms of elementary functions.
We'll have to leave it in its current form.

STEP 8

Let's **substitute** x1x - 1 back in for uu to get our answer in terms of xx: 3(x1+1)12x1dx=3x12x1dx\int \frac{3(x-1 + 1)^{-\frac{1}{2}}}{x-1} \, dx = \int \frac{3x^{-\frac{1}{2}}}{x-1} \, dx

STEP 9

Let's try another substitution.
Let u=xu = \sqrt{x}, so x=u2x = u^2 and dx=2ududx = 2u \, du.
Substituting, we get 3x3/2xdx=3u3udx=3u(u21)(2udu)=61u21du\int \frac{3}{x^{3/2} - \sqrt{x}} \, dx = \int \frac{3}{u^3 - u} \, dx = \int \frac{3}{u(u^2 - 1)} (2u \, du) = 6 \int \frac{1}{u^2 - 1} \, du

STEP 10

We know that 1u21du=12lnu1u+1+C\int \frac{1}{u^2 - 1} \, du = \frac{1}{2} \ln \left| \frac{u-1}{u+1} \right| + C So, our integral becomes 61u21du=3lnu1u+1+C6 \int \frac{1}{u^2 - 1} \, du = 3 \ln \left| \frac{u-1}{u+1} \right| + C

STEP 11

Substituting back u=xu = \sqrt{x}, we get 3lnx1x+1+C3 \ln \left| \frac{\sqrt{x}-1}{\sqrt{x}+1} \right| + C

STEP 12

Our final answer is 3lnx1x+1+C3 \ln \left| \frac{\sqrt{x}-1}{\sqrt{x}+1} \right| + 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