Integral of a Fraction: Solving with Substitution and Integration by Parts

[aquitaine]

Homework Equations

\int\sqrt{x^{3}}+1/\sqrt{x}+1

The Attempt at a Solution

I tried substitution, but that wouldn't work. I tried integration by parts, but I must not have done it properly since my answer was several factors of x off. Am I missing something?

 

[rocomath]

Use parenthesis! I'm assuming your problem is:

\int\frac{\sqrt{x^3} +1}{\sqrt x+1}dx

So, where is your work? Thanks :)

 

[dirk_mec1]

Homework Equations

\int\sqrt{x^{3}}+1/\sqrt{x}+1

The Attempt at a Solution

I tried substitution, but that wouldn't work. I tried integration by parts, but I must not have done it properly since my answer was several factors of x off. Am I missing something?

Please use brackets!

Now it isn't clear what your integral is.

Is it the integral below? Is het root taken only of x^3 and x or the whole term?


\int \frac{ \sqrt{x^3}+1}{\sqrt{x}+1} \mbox{d}x

 

[RTW69]

Assuming rocomath is correct in formulating the problem, try long division before integration

 

[aquitaine]

Use parenthesis! I'm assuming your problem is:

\int\frac{\sqrt{x^3} +1}{\sqrt x+1}dx

So, where is your work? Thanks :)

Right. Here's my work:

u=x^{3/2}+ 1 du= \frac{3}{2}x^{1/2}
dv = x^{1/2} + 1 v= 2x^{3/2} + x


After that I decided to check the answer before proceeding further to see if I was on the right track, which I wasn't. The books answer was (1/2)x^2 - (3/2)x\sqrt{x} + x + C, and I'm at a loss as to how that happened.

 

[rocomath]

I let x=u^2 \rightarrow dx=2udu

Then I used polynomial division.

 

[RTW69]

divide sqrt(x)+1 into x*sqrt(x)+1, the integration then becomes very straight forward

 

[dynamicsolo]

divide sqrt(x)+1 into x*sqrt(x)+1, the integration then becomes very straight forward

Quite so, or factor x^{3/2} + 1 as a sum of two cubes...

 

[rocomath]

Quite so, or factor x^{3/2} + 1 as a sum of two cubes...

Oh! Very nice, didn't even notice that.

 

[aquitaine]

thanks guys!