Question

Integration

Answer 1

Try substituting,

\[\large{\tan{\theta} = u}\]

Answer 2

Okay. wait =)

Answer 3

Take your time :)

Answer 4

got ln tan theta +c =)

Answer 5

how about this?

Answer 6

Well first of all see that the degree of polynomial in numerator = 3 and in denominator = 2

Whenever, the degree of polynomial in numerator > denominator, I generally, solve the numerator in such a way that it becomes something easier to tackle. For example, here,

\[\large{x^3+3x = x(x^2+1) + 2x}\]

Now, by doing this see what I can get,

\[\large{I = \int\cfrac{x(x^2+1)+2x}{x^2+1}dx}\]

\[\large{=\int xdx +2\int\cfrac{x}{x^2+1}dx}\]

Answer 7

Tell me if you have any doubt till now.

Answer 8

Any doubt @silverxx ?

Answer 9

still digesting it haha =)

Answer 10

can we also u sub here?

Answer 11

Take your time.

Answer 12

*use u sub here?

Answer 13

Gocha. x^2/2 + ln x^2 + 1 +c. the answer

Answer 14

Thank you so much @vishweshshrimali5

Answer 15

No problem and great work.

Answer 16

wait. can i ask again? having doubts to this problem.

Answer 17

the answer is 1/2 (lnx)^2 + c. have no idea to the 1/2 thing

Answer 18

Okay lets see

Answer 19

If you can get (lnx)^2 part then very good.. You have got the main part. So, I am just going to post the solution (incomplete):

\[\large{I = \int\cfrac{\ln x}{x}dx}\]

Substitute, \(\large{\ln x = u}\)

\[\large{\implies du = \cfrac{1}{x}dx}\]

So, your integral becomes,

\[\large{I = u*du}\]

Now, what will be I ?

Answer 20

Ohh sorry in the last step its actually:

\[\large{I = \int udu}\]

Answer 21

Yes that was my last step and then im lost

Answer 22

Ok lets see:

\[\large{I' = \int x^n dx}\]
\[\large{I' = \cfrac{x^{n+1}}{n+1}}\]

Now, what would be:

\[\large{I = \int u du}\]

Answer 23

Here, n = 1.

Answer 24

okay.. okay! Gets!! =)

Answer 25

Thank you

Answer 26

No problem ! :)