Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


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}$

share|cite|improve this question
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
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 $u = 2x^3 + 5$.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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 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 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 Dec 10 '12 at 19:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.