Question

Find the integration

Answer 1

\[(3x+4)\sqrt{3x-4} \] dx

Answer 2

my first instinct is to integrate by parts with\[u=3x+4\\dv=\sqrt{3x-4}dx\]do you know how to do that?

Answer 3

I believe , I wasn't taught how to integrate in parts.

Answer 4

then I will try to find another way

Answer 5

but just to get idea, how do you integrate in parts ?

Answer 6

\[\int udv=uv-\int vdu\\\int(3x+4)\sqrt{3x-4}dx=(3x+4)\frac13(3x-4)^{3/2}\cdot\frac23-\int(3x-4)^{3/2}dx\]

Answer 7

Doesn't seem easy. By the way , the book , I found the problem in, has it "half-solved" only the final part. Will seeing the answer help you ?

Answer 8

maybe

Answer 9

u substitution might work

Answer 10

yeah but what sub? I am drawing a blank

Answer 11

\(u = \sqrt{3x-4} \implies du = \dfrac{3}{2\sqrt{3x-4}}dx = \dfrac{3}{2u}dx \implies dx = \dfrac{2udu}{3}\)

Answer 12

thank you @rational I am rusty :P

Answer 13

3x(3x -4)^1/2 + 4(3x -4)^1/2 = [3x +4 -4](3x-4)^1/2 + 4(3x-4)^1/2 =
(3x -4)(3x-4)^1/2 + 4(3x-4)^1/2 + 4(3x-4)^1/2
= (3x -4) + 8(3x -4 )^1/2 then by integrating with respect to x

Answer 14

that looks reasonable as well

Answer 15

I don't understand a word.

Answer 16

what do you mean?

Answer 17

last line should be
(3x -4)^3/2 + 8(3x -4 )^1/2

Answer 18

The integration was (3x+4)(3x-4)^1/2 , well how did the book writer turned it into that

Answer 19

distribute (3x+4)

Answer 20

let u=(3x-4)^1/2
(3x+4)u=3xu+4u=3x(3x-4)^1/2+4(3x-4)^1/2

Answer 21

Now , I see.

Answer 22

Thanks so much guys.

Answer 23

welcome!

Answer 24

I got last question.

Answer 25

ok i;ll find it