Integration problem

[Confuzzled :/]

Hi please could someone help me solve the integral ∫ x(x^2+1)^1/2 dx.
I have tried to solve it using integration by parts but got the wrong answer:
u = x
dv/dx = (x^2+1)^1/2
du/ dt = 1
v = 2/3(x^2 + 1)^3/2

2/3x(x^2 + 1)^3/2 - ∫2/3(x^2 + 1)^3/2 dx
2/3x(x^2 + 1)^3/2 - 2/3[ 2/5(x^2 +1)^5/2] +c
2/3 x (x^2 + 1)^3/2 - 4/15 (x^2 +1)^5/2 + c

Thank you

 

[MarkFL]

Hello, and welcome to FMH!

I would use a simple \(u\)-substitution:

[MATH]u=x^2+1\implies du=2x\,dx[/MATH]
And we have:

[MATH]I=\frac{1}{2}\int u^{\frac{1}{2}}\,du[/MATH]
Can you proceed?

 

[Dr.Peterson]

Hi please could someone help me solve the integral ∫ x(x^2+1)^1/2 dx.
I have tried to solve it using integration by parts but got the wrong answer:
u = x
dv/dx = (x^2+1)^1/2
du/ dt = 1
v = 2/3(x^2 + 1)^3/2

2/3x(x^2 + 1)^3/2 - ∫2/3(x^2 + 1)^3/2 dx
2/3x(x^2 + 1)^3/2 - 2/3[ 2/5(x^2 +1)^5/2] +c
2/3 x (x^2 + 1)^3/2 - 4/15 (x^2 +1)^5/2 + c

Your v is incorrect. To check, try differentiating it. (In fact, doing that might remind you to use substitution!)

If you try to pursue this method by correcting that, you'll find it much more difficult, I think.

 

[Confuzzled :/]

Hello, and welcome to FMH!

I would use a simple \(u\)-substitution:

[MATH]u=x^2+1\implies du=2x\,dx[/MATH]
And we have:

[MATH]I=\frac{1}{2}\int u^{\frac{1}{2}}\,du[/MATH]
Can you proceed?

Thanks for your help! I've got the right answer now
I = (1/2 )(2/3)u^3/2 +c
I = 1/3u^3/2 +c
I = 1/3(x^2 +1)^3/2 + c

 

[Confuzzled :/]

Your v is incorrect. To check, try differentiating it. (In fact, doing that might remind you to use substitution!)

If you try to pursue this method by correcting that, you'll find it much more difficult, I think.

Thanks for your reply!
I got dv/dx = 2x(x^2+1)^1/2 so I must have integrated it incorrectly.
I've now managed to get the right answer using substitution.