# Math Help - Difficult Integral

1. ## Difficult Integral

Hi all,

Im studying calculus now and Ive come across this question in a workbook and im struggling with it. Its to find this integral:

∫ 7x ² / x ³ + 1 dx

If someone could show me it would be a great help and much of a relief.

Mitch

2. Originally Posted by Mitch281008
Hi all,

Im studying calculus now and Ive come across this question in a workbook and im struggling with it. Its to find this integral:

∫ 7x ² / x ³ + 1 dx

If someone could show me it would be a great help and much of a relief.

Mitch
Make the substitition $u=x^3+1 \implies du=3x^2dx$

This gives

$\int \frac{7x^2}{x^3+1}dx=\frac{7}{3}\int\frac{1}{u}du$

Now just integrate from here

3. Have you used U substitution?

Try setting $u=x^3$

Then: $du=3x^2dx$

-> $\frac{7du}{3} = 7x^2$

The integral then becomes: $\frac{7}{3}\int{\frac{1}{u}}du +\int1dx$

4. Originally Posted by Riyzar
Have you used U substitution?

Try setting $u=x^3$

Then: $du=3x^2dx$

-> $\frac{7du}{3} = 7x^2$

The integral then becomes: $\frac{7}{3}\int{\frac{1}{u}}du +\int1dx$
You are supposing $u = x^3$

NOTE THAT you have to suppose

$u = {x^3}+ 1$
then, $du = 3x^2 dx$

$\frac{7}{3}\int{\frac{1}{u}}du$

= $\frac{7}{3} log(u) + C$

substitute $u = {x^3}+ 1$

and you're are smart enough to write the answer

5. Ok thanks guys.

It was one of those questions where i could not get an answer no matter what i tried.

i appreciate the help

cheers,
Mitch

6. Ok I have another problem, this one ive solved but i just want to check the answer to see if ive got it.

∫ 5e^(3x+1) = 5 (e^(3x+1)) / (3x+1) + c

i hope thats clear, its annoying not being able to make fractions look like fractions lol.

thanks for the support,

Mitch

7. Originally Posted by Mitch281008
Ok I have another problem, this one ive solved but i just want to check the answer to see if ive got it.

∫ 5e^(3x+1) = 5 (e^(3x+1)) / (3x+1) + c

i hope thats clear, its annoying not being able to make fractions look like fractions lol.

thanks for the support,

Mitch

$\int 5e^{3x+1} dx$

let $u = 3x+1$

then $du = 3dx$

so, $dx = \frac{du}{3}$

you have,

$\int \frac{5e^{u}du}{3}$

= $\frac{5e^u}{3}+C$

substitute $u = 3x+1$

you get,

$\frac{5e^{3x+1}}{3}+C$