# Integration by subsitution?

#### cwashburn

I am trying to integrate this by parts and not making progress. Is this best done with substitution?

$$\displaystyle \int_1^3 \frac{x}{(x+1)^5}$$

#### Ackbeet

MHF Hall of Honor
I certainly would. Basically, it allows you to put the variable shift x+1 into the numerator, where it is not raised to the 5th power.

#### cwashburn

I certainly would. Basically, it allows you to put the variable shift x+1 into the numerator, where it is not raised to the 5th power.
So you're stating that u substitutes for $$\displaystyle (x+1)^{-5}$$ and I integrate from there and subsitute back at the end?

#### Ackbeet

MHF Hall of Honor
I would do u = x+1.

#### Also sprach Zarathustra

By substitution

$$\displaystyle \int\frac{x}{(x+1)^5}dx = ?$$

Let $$\displaystyle x+1=t$$, hence, $$\displaystyle dx=dt$$ and $$\displaystyle x=t-1$$

Now, $$\displaystyle \int\frac{x}{(x+1)^5}dx=\int\frac{t-1}{t^5}dt=\int\frac{t}{t^5}dt-\int\frac{1}{t^5}dt=-\frac{t^{-3}}{3}+\frac{t^{-4}}{4}+C$$

And now, back to origins, we get:

$$\displaystyle \int\frac{x}{(x+1)^5}dx=-\frac{(x+1)^{-3}}{3}+\frac{(x+1)^{-4}}{4}+C$$

#### Ackbeet

MHF Hall of Honor
Your original integral had limits. If you have limits, there's no need to get back to the original variable of integration, so long as you transform the limits into the new variable. Also, I think your integrations are incorrect. Don't skip so many steps!

#### Ackbeet

MHF Hall of Honor
Oh, I must have been seeing different numbers in the denominator. It's been a long day!

Still need to plug in limits, though. Perhaps we could let cwashburn do that?

#### TheCoffeeMachine

Perhaps we could let cwashburn do that?
I guess so. It's always good to leave the OP something to do. (Evilgrin)

#### cwashburn

Per your suggestion Ackbeet I substituted y for x+1.

That led me to x = y-1 and so I had $$\displaystyle (y-1)*y^{-5}$$. Multiplying through I get $$\displaystyle \int y^{-4} - \int y^{-5}$$.

So after integrating I have $$\displaystyle \int -\frac{1}{3y^{3}} - \int-\frac{1}{4y^{4}}$$

I've been balancing with wolframalpha against the result I set up as a formula in Excel.

Interestingly my integration $$\displaystyle \frac{-1}{3y^{3}}\mid_1^3 - \frac{-1}{4y^{4}}\mid_1^3$$ works, but on the interval of 2 to 4, versus 1 to 3, which are the original boundaries.

Do I need to increase the intervals by 1 because prior to substitution I had x+1? Is that why 2 to 4 works out to the right answer on $$\displaystyle \frac{-1}{3y^{3}}\mid_{y+1}^{x+1} - \frac{-1}{4y^{4}}\mid_{y+1}^{x+1}$$ where x and y are my original limits?