How do you work out $\int \ln(x) dx$? (I want two methods.)

A

Parts ($u = \ln(x),~v'=1$)

Hint:

Yep, it drops out nicely.

B

You look it up - it's $\frac{1}{x} + C$

Hint:

No, that's differentiating

C

You can only do it numerically

Hint:

Nut-uh. You can do it numerically, of course, but it's not the only way.

D

Substitution: $x = e^u$

Hint:

Not a popular way, but a good way.

E

Parts ($u = 1,~v'=\ln(x)$)

Hint:

No, if you knew how to integrate $ln$, you wouldn't be in this mess.

Question 5

How do you work out $\int \tan^2(x) dx$?

A

Substitution: start from $\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$

Hint:

It's a nice thought, but no. You end up with a $\frac{\tan(2x)}{\tan(x)}$ to integrate, which is no good at all.

B

You can only do it numerically.

Hint:

Our survey said: ee-aw.

C

Trig identity: $\tan^2(x) = \sec^2(x) - 1$

Hint:

$sec^2(x)$ integrates to $\tan(x)$. It's in the book.

D

Parts: $u = \tan(x),~v'=\tan(x)$

Hint:

Good luck with integrating $\sec^2(x) \ln(\sec(x)\tan(x))$ in the second step.

E

Parts: $u = \tan^2(x),~v'=1$

Hint:

This ends up as $x\tan^2(x) - \int 2x \tan^2(x)sec(x) dx$. I reckon it's possible, but I don't fancy it.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
Get Results

There are 5 questions to complete.

←

List

→

Return

Shaded items are complete.

1

2

3

4

5

End

Return

You have completed

questions

question

Your score is

Correct

Wrong

Partial-Credit

You have not finished your quiz. If you leave this page, your progress will be lost.

Correct Answer

You Selected

Not Attempted

Final Score on Quiz

Attempted Questions Correct

Attempted Questions Wrong

Questions Not Attempted

Total Questions on Quiz

Question Details

Results

Date

Score

Hint

Time allowed

minutes

seconds

Time used

Answer Choice(s) Selected

Question Text

All done

Need more practice!

Keep trying!

Not bad!

Good work!

Perfect!

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

Maybe better if you show the answer for the actual maths at the end.

Just telling you but, isn’t the v=1 on the ln(x) question suppose to be v’=1? (2^x has one choice where it is v’=1)
Also, there is 80% chance that “non of the above” is not at the bottom so shouldn’t it be non of the others?

## MathbloggingAll

C4 Integration Quiz (tough stuff) http://t.co/NFQ24eRzbb

## A

Good questions thanks

Maybe better if you show the answer for the actual maths at the end.

Just telling you but, isn’t the v=1 on the ln(x) question suppose to be v’=1? (2^x has one choice where it is v’=1)

Also, there is 80% chance that “non of the above” is not at the bottom so shouldn’t it be non of the others?

## Colin

Thanks for the feedback — I’ve shied away from giving the answers to the integrals, mainly out of laziness.

Thanks especially for the corrections, I’ve fixed them 🙂

## mkami wambura

where are questions?????

## Colin

They should show up after you click the big blue ‘Start’ button.

## aayush ganesh

Differentiating log base 2 is actually fairly straightforward.

y = log2(x)

2^y = x

dx/dy = 2^y ln(2)

dy/dx = 1/((2^y)ln(2))

dy/dx = 1/xln(2)

Imo, this is easier than writing 2^x in terms of e^x and following the e^x integration by recognition rules.

The only reason it’s not used is because it’s not on the syllabus 🙁

They put differentiating and integrating any exponential but not any logarithm. Doesn’t make much sense to me.

## Colin

I’d probably do it as:

$y = \log_2(x) = \frac{\ln(x)}{\ln(2)}$

$\dydx = \frac{1}{x} \times \frac{1}{\ln(2)}$ directly.