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I understand the basics of anti-derivatives, and have been given the following questions:

1: $ \int [t^3 - 5e^t - 4\cos(t)] dt $

My answer:

$ \frac{t^4}{4} - e^{5t} + \sin(4t) + c $

2: $ \int [\frac{8}{x} + 5\sec(x)\tan(x)] dx$

My answer:

$8\ln(x) + \sec^2(5x) + c$

3: $ \int \frac{4x}{(x^2 +5)^3} dx $

My answer:

None yet, slightly confused since it's a quotient.

4: $ \int [x^2 \sqrt{2x^2 + 5}] dx $

My answer:

None yet, confused as well.

So I'm wondering if someone can check number 1 & 2, and then explain/show 3 & 4?

EDIT

These are what I get when I redo them (I've done my best to just look at suggestions rather than answers):

1: $ \frac{t^4}{4} - 5e^t - 4sin(t) + c $

2: $ 8ln(x) + 5sec(x) + c $

3: $ 2 \frac{(x^2 + 5)^{-2}}{-2}$

4: $ \frac{1}{6} \frac{(2x^3 + 5)^{3/2}}{3/2}$

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The first one is wrong. The second has a small error. You can check the result of an integration yourself, by differentiating. For the third, let $u=x^2+5$. –  André Nicolas Dec 10 '12 at 18:00
    
Yeah. 1 is wrong. It should be $\frac{t^4}{4}-5e^{t}-4\sin t +c$. –  juniven Dec 10 '12 at 18:04
    
@AndréNicolas In 1. is the issue $-e^5t$? As for 2. Is it $sec(5x)$ instead of $sec^2(5x)$ –  StrugglingWithMath Dec 10 '12 at 18:05
    
The integral of $\sec x\tan x$ is $\sec x$. –  André Nicolas Dec 10 '12 at 18:08
    
So the answer in 2 is $8\ln x+5\sec x +c$ –  juniven Dec 10 '12 at 18:15
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2 Answers

up vote 3 down vote accepted

$1$ and $2$ are not quite right. You seem to understand the idea but are missing one thing:

The derivative of for example $\sin(4t)$ by the chain rule would be $4 \cos(4t)$ not $4 \cos(t)$. This should show you why you shouldn't bring the constant inside the function you're taking the antiderivative of.

In general if you have a constant multiplied by a function you should take the antiderivative as if the constant were not there, and then multiply that by the constant.

For example for $4 \cos(t)$, we'll ignore the $4$ and take the antiderivative of $\cos(t)$ which is $\sin(t)$, then multiply it by $4$ to get $$ 4 \sin(t).$$ To check that this will work you should take the derivative and make sure its the what you want it to be!

You have made this error a couple if times in the first two and I advice you to try and find all the places you have and fix it.

For problem $3$ I advise you to try the $u$-sub $ u = x^2 + 5$.

And for problem $4$ I think the integral is intended to be $x^2\sqrt{2x^3 + 5}$ in which case I would suggest the $u$-sub $u = 2x^3 + 5$.

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For problem 4. If u = $2x^3 + 5$, then du = $6x^2$. How do I deal with du when plugging it in? –  StrugglingWithMath Dec 10 '12 at 18:30
    
@StrugglingWithMath Yes you are right it will be $\frac{1}{6}\mathrm{d}u$! You have $\mathrm{d}u = 6x^2 \mathrm{d}x$, so $\frac{1}{6} \mathrm{d}u = x^2 \mathrm{d}x$, you see that $x^2 \mathrm{d}x$ is in the integral, so replace that with $\frac{1}{6} \mathrm{d}u$! –  Deven Ware Dec 10 '12 at 18:36
    
So: $\frac{1}{6}du * u^{1/2}$ which is $ \frac{1}{6} \frac{u^{3/2}}{3/2}$ then plug $2x^3+5$ back in for u? –  StrugglingWithMath Dec 10 '12 at 18:41
    
@StrugglingWithMath yep thats right! –  Deven Ware Dec 10 '12 at 18:51
    
Thank you so much! :) –  StrugglingWithMath Dec 10 '12 at 18:57
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First of all, you can always check whether an antiderivative is right by differentiating and checking whether you get what you started with.

For 1: Remember that $\int a f(x) \ dx = a \int f(x) \ dx$, that $\int e^x \ dx = e^x + C$ and that $(\sin x)' = \cos x$, not $-\cos x$. Also look at the derivative of $\sin(4x)$ and compare it with the derivative of $4 \sin x$.

For 2: What's the derivative of $\sec^2 x$, and what's the derivative of $\sec x$?

For 3: Try a substitution $u = x^2 + 5$.

For 4: Are you sure that's what you're supposed to do? That integral is pretty difficult, certainly not what I'd expect someone who's just starting with this to be able to do.

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Similar to the question I gave Deven, but for problem 3. How do I deal with the du part since it doesn't equal what is left in the equation with u substituted in? du = 2x and there is a 4x in the original equation. –  StrugglingWithMath Dec 10 '12 at 18:31
    
@StrugglingWithMath: since $u = x^2 + 5$, we get that $du = 2x\ dx$. But in your integral you have $4x\ dx$. Well, no problem! $4x\ dx = 2\times 2x\ dx = 2\ du$. –  Javier Badia Dec 10 '12 at 18:47
    
@StrugglingWithMath: Also, your first integral is not quite right yet, you have a sign issue. You get that the antiderivative of $-4\cos x$ is $4 \sin x$. Well, try differentiating $4\sin x$. What do you get? –  Javier Badia Dec 10 '12 at 18:49
    
#1: 4 sinx -> 4 cosx. I'll change the sign :) as for #3: $2 u^-3$ then $2 \frac{u^{-2}}{-2}$ finally $2 \frac{(x^2 + 5)^{-2}}{-2}$ Is that correct? –  StrugglingWithMath Dec 10 '12 at 18:55
    
@StrugglingWithMath: Yes, that's right. Always remember, you can check yourself: just differentiate. –  Javier Badia Dec 10 '12 at 19:11
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