Question

solve this
∫ 1/(√(5x+2) + √(2x+1) ) dx

Answer 1

No, it's all under the denominator, mihirb.

Answer 2

oh it is didnt notice that

Answer 3

yes its all in denominator

Answer 4

$$\int \frac{1}{\sqrt {5x + 2} + \sqrt {2x + 1}} \, dx$$

Answer 5

oh srry

Answer 6

yeah thats the question

Answer 7

It's a little bit weird, but u-substitution works.

Answer 8

yup

Answer 9

i tried but i m not getting it..

Answer 10

First, rationalize the integrand by multiplying by the conjugate of the denominator. Can you do that?

Answer 11

yeah

Answer 12

i m getting after rationalizing\[\frac{ \sqrt{5x+2} +\sqrt{2x+1} }{ 3x+3 }\]

Answer 13

sorry there should be -ve sign in numerator

Answer 14

Yes, now use partial fractions to split up the integral.

Answer 15

done.. then?

Answer 16

and 3x+1 in the denominator

Answer 17

No, it's -3x -1.

Answer 18

what i did't understand?

Answer 19

multiplying by conjugate in the denominator should lead to
5x+2-(2x+1)
which is 3x+1 if you distribute the negative properly

Answer 20

Which is just the same thing as a minus sign over the whole thing, which is a minus sign in the denominator. Er, well, this is what I'm using, so go with this for clarity, it's just a sign mix-up:
$$\int \frac {\sqrt {2x + 1}} {-3x - 1}\, dx - \int \frac {\sqrt {5x + 2}} {-3x - 1}\, dx$$

Answer 21

what after this\[\frac{ \sqrt{5x+2} }{ 3x+3 } - \frac{ \sqrt{2x+1} }{ 3x+3 }\]

Answer 22

okay then?

Answer 23

Substitute \(u = 5x + 2\) and its corresponding \(du = \frac{5}{2\sqrt {5x + 2}}\).

Answer 24

We're starting with that integral because it's a bit easier

Answer 25

if u=5x+2 then how u got the value of du?

Answer 26

Sorry, I meant \(u = \sqrt {5x + 2}; du = \frac {5}{2\sqrt{5x + 2}} dx\).

Answer 27

okay .. now i got it

Answer 28

then?

Answer 29

We have:
$$-\frac{2}{5} \int \frac{u^2}{-\frac 3 5 (u^2 - 2) - 1} \, du$$
Now, just cancel out stuff with a bit of interesting manipulation.
$$-\frac{2}{5} \int -\frac{5u^2}{3u^2 - 1} \, du$$

Answer 30

Now factor out all of those constants. Move them to other integral if you please.

Answer 31

What should i do now?? and what abt the other term?

Answer 32

@dan815 u got it wrong.. the whole expression is in denominator..

Answer 33

see it in one the comments..

Answer 34

oh ok

Answer 35

like this?