Integration by parts - find the mistake

[Glathberif]

Hello,

I have tried to solve the following integral and get a result which is wrong (because when I derivate it, I get something different).

Let int( ) be the integral sign. Plus, I'll skip dx and C to help readability.

int(x⁵*sqrt(x³+1))

Let u=x⁵ and dv=sqrt(x³+1)
So, du=5x⁴

Now, let's find v.

Let t=x³+1

v=int(sqrt(t))
=2/3*t^3/2
=2/3*(x³+1)^3/2

int(x⁵*sqrt(x³+1))
=int(udv)
=uv-int(vdu)
=2/3*x⁵*(x³+1)^3/2-10/3*int(x⁴*(x³+1)^3/2

Now, you can see the degree of x in the new integral is smaller, so I basically did it again and again until it disappeared, but when I derivate it, it just doesn't fit, so I guess I made the same mistake in each step, but can't find it (probably the method itself is wrong?)

Here is the derivative if it can help: http://www.wolframalpha.com/input/?i=d%282%2F3x^5*%28x^3%2B1%29^%283%2F2%29%29%2Fdx-10%2F3*%28x^4*%28x^3%2B1%29^%283%2F2%29%29 (look at the alternative form, you can see the square root is correct but not the degree nor the scalar of x)

Thanks for your help!

 

[HallsofIvy]

I would NOT use integration by parts because, as you saw, the integrand gets more complicated. Instead write the integral as \(\displaystyle \int x^5\sqrt{x^3+ 1}dx= \int x^3\sqrt{x^3+ 1}(x^2dx)\) and use the substitution \(\displaystyle u= x^3+ 1\). Then \(\displaystyle \sqrt{x^3+ 1}\), \(\displaystyle x^2= u- 1\) and \(\displaystyle x^2dx= \frac{1}{3}du\)
The integral becomes \(\displaystyle \int (u- 1)\sqrt{u}(\frac{1}{3}du)= \frac{1}{3}\int u^{3/2}- u^{1/2} du\)

 

[Glathberif]

I looked at the solution and it indeed said that we should do it another way, but that's not my problem. My problem is that I've made a mistake and can't seem to find it.

 

[lookagain]

nstead write the integral as . . . \int x^3\sqrt{x^3+ 1}(x^2dx)[/tex] and use the substitution \(\displaystyle u= x^3+ 1\). > > > Then \(\displaystyle \sqrt{x^3+ 1}\), \(\displaystyle x^2= u- 1\) < < < and \(\displaystyle x^2dx= \frac{1}{3}du\)

\(\displaystyle \ \ \ \ \ \ \ \)There were a couple of accidental errors above. \(\displaystyle \ \ \) Corrections:
\(\displaystyle . . . \int x^3\sqrt{x^3+ 1}(x^2dx)\) and use the substitution \(\displaystyle u= x^3+ 1 \ \). Then \(\displaystyle \sqrt{x^3+ 1} = \sqrt{u} \ \), \(\displaystyle \ x^3= u- 1 \ \) and \(\displaystyle \ x^2dx = \frac{1}{3}du\)

 

[Deleted member 4993]

Hello,

I have tried to solve the following integral and get a result which is wrong (because when I derivate it, I get something different).

Let int( ) be the integral sign. Plus, I'll skip dx and C to help readability.

int(x⁵*sqrt(x³+1))

Let u=x⁵ and dv=sqrt(x³+1)
So, du=5x⁴

Now, let's find v.

Let t=x³+1 → x³ = t-1 → x = (t-1)^(1/3) → dx = 1/3 (t-1)^(-2/3) dt

v=int(sqrt(t))
=2/3*t^3/2
=2/3*(x³+1)^3/2

This is the origin of the problem and it stems from the fact that you have dropped 'dx' from your expressions
It should be:

v = \(\displaystyle \displaystyle\frac{1}{3}\int \sqrt{t}\frac{1}{(t-1)^{\frac{2}{3}}} dt \)


int(x⁵*sqrt(x³+1))
=int(udv)
=uv-int(vdu)
=2/3*x⁵*(x³+1)^3/2-10/3*int(x⁴*(x³+1)^3/2

Now, you can see the degree of x in the new integral is smaller, so I basically did it again and again until it disappeared, but when I derivate it, it just doesn't fit, so I guess I made the same mistake in each step, but can't find it (probably the method itself is wrong?)

Here is the derivative if it can help: http://www.wolframalpha.com/input/?i=d%282%2F3x^5*%28x^3%2B1%29^%283%2F2%29%29%2Fdx-10%2F3*%28x^4*%28x^3%2B1%29^%283%2F2%29%29 (look at the alternative form, you can see the square root is correct but not the degree nor the scalar of x)

Thanks for your help!

Moral of the story - do not drop 'dx' - even if it looks ugly

 

[Glathberif]

Moral of the story - do not drop 'dx' - even if it looks ugly

Thanks! I knew there was something wrong with my change of variable. I may need to work on that again.