Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

(x+1)^3/(x^2)dx

(x+1)/(x^2) dx

1/(1+4(x^2)) dx

x(x-10)^10 dx

(cosx)/(1+(sinx)^2) dx

(1+tanx)^3/(cosx)^2

Thanks :)

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- Jan 7th 2010, 01:34 PMali2141U-Substitution Integrals
Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

(x+1)^3/(x^2)dx

(x+1)/(x^2) dx

1/(1+4(x^2)) dx

x(x-10)^10 dx

(cosx)/(1+(sinx)^2) dx

(1+tanx)^3/(cosx)^2

Thanks :) - Jan 7th 2010, 01:40 PMDrexel28
- Jan 7th 2010, 01:42 PMpickslides
- Jan 7th 2010, 01:43 PMadkinsjr
It's not clear what you did here. If you want to integrate $\displaystyle \int\frac{(x+1)^3}{x^2}dx$, you can just expand the numerator $\displaystyle \int\frac{x^3+3x^2+3x+1}{x^2}dx$ and then split it up into a sum. Then you should be able to apply the general formula for polynomial integrands.

- Jan 7th 2010, 01:56 PMali2141
I was trying to use u-substitution.

Wait, so you mean do: x^3/x^2 + 3x^2/x^2 + 3x/x^2 + 1/x^2 ?

What would be the answer to that then? - Jan 7th 2010, 02:12 PMadkinsjr
You don't see how to integrate that? If you're studying the substitution rule, I assume you must be familiar with the basic methods of integral calculus. Just use the sum rule, and integrate each term using the formula$\displaystyle \int ax^ndx=\frac{a}{n+1}x^{n+1}+C$

- Jan 7th 2010, 02:20 PMali2141
No, I get that if you have x^3, the antiderivative would be x^4/4

But I don't know how that would apply to something like x^3/x^2 - Jan 7th 2010, 02:24 PMpickslides
- Jan 7th 2010, 02:26 PMali2141
oh wow. (Headbang)

okay, i feel like an idiot now. - Jan 7th 2010, 03:22 PMTWiX
- Jan 7th 2010, 03:39 PMProve It
The first two have been taken care of.

for $\displaystyle \int{\frac{1}{1 + 4x^2}\,dx}$ you do not use a $\displaystyle u$ substitution. You use a trigonometric substitution.

$\displaystyle \int{\frac{1}{1 + 4x^2}\,dx} = \int{\frac{1}{4\left(\frac{1}{4} + x^2\right)}\,dx}$

$\displaystyle = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx}$.

Now use the substitution $\displaystyle x = \frac{1}{2}\tan{\theta}$.

Note that $\displaystyle \frac{dx}{d\theta} = \frac{1}{2}\sec^2{\theta}$.

So $\displaystyle dx = \frac{1}{2}\sec^2{\theta}\,d\theta$.

Therefore

$\displaystyle = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx} = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + \left(\frac{1}{2}\tan{\theta}\right)^2}\,\frac{1}{ 2}\sec^2{\theta}\,d\theta}$

$\displaystyle = \frac{1}{8}\int{\frac{1}{\frac{1}{4}(1 + \tan^2{\theta})}\,\sec^2{\theta}\,d\theta}$

$\displaystyle = \frac{1}{2}\int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$

$\displaystyle = \frac{1}{2}\int{\frac{1}{\sec^2{\theta}}\,\sec^2{\ theta}\,d\theta}$

$\displaystyle = \frac{1}{2}\int{1\,d\theta}$

$\displaystyle = \frac{1}{2}\theta + C$.

Now remembering that $\displaystyle x = \frac{1}{2}\tan{\theta}$

$\displaystyle 2x = \tan{\theta}$

$\displaystyle \theta = \arctan{2x}$.

So finally our answer is $\displaystyle \frac{1}{2}\arctan{2x} + C$. - Jan 7th 2010, 03:46 PMTWiX
- Jan 7th 2010, 03:48 PMProve It
For $\displaystyle \int{x(x - 10)^{10}\,dx}$ you could expand everything, but that could get messy.

Instead, integration by parts will need to be used.

Let $\displaystyle u = x$ so that $\displaystyle du = 1$.

Let $\displaystyle dv = (x - 10)^{10}$ so that $\displaystyle v = \int{(x - 10)^{10}\,dx} = \frac{1}{11}(x - 10)^{11}$.

Therefore $\displaystyle \int{x(x - 10)^{10}\,dx} = \frac{1}{11}x(x - 10)^{11} - \int{\frac{1}{11}(x - 10)^{11}\,dx}$

$\displaystyle = \frac{1}{11}x(x - 10)^{11} - \frac{1}{132}(x - 10)^{12} + C$

$\displaystyle = \frac{1}{11}(x - 10)^{11}\left[x - \frac{1}{12}(x - 10)\right] + C$

$\displaystyle = \frac{1}{11}(x - 10)^{11}\left(x - \frac{1}{12}x + \frac{5}{6}\right)$

$\displaystyle = \frac{1}{11}(x - 10)^{11}\left(\frac{11}{12}x + \frac{5}{6}\right)$

$\displaystyle = \frac{1}{132}(x - 10)^{11}(11x + 10)$. - Jan 7th 2010, 03:51 PMProve It
For $\displaystyle \int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx}$, note that

$\displaystyle \frac{1}{\cos^2{x}} = \sec^2{x}$.

So $\displaystyle \int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})\sec^2{x}\,dx}$

Now let $\displaystyle u = 1 + \tan{x}$ so that $\displaystyle \frac{du}{dx} = \sec^2{x}$.

The integral becomes

$\displaystyle \int{u\,\frac{du}{dx}\,dx}$

$\displaystyle = \int{u\,du}$

$\displaystyle = \frac{1}{2}u^2 + C$

$\displaystyle = \frac{1}{2}(1 + \tan{x})^2 + C$. - Jan 8th 2010, 09:10 AMroshanhero
I am sorry but shouldnot there have been

$\displaystyle

\int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})^3\sec^2{x}\,dx}

$