Question

Integration

Answer 1

\[\int\limits\frac{ 5dx }{ 2x^2+7x-4 }\]

Answer 2

fIRST STEP, TAKEOUT THE 5 FROM THE INTEGRAL


caps...

Answer 3

Ha, caps

Answer 4

\[\large 5 \int\limits \frac{dx}{2x^2+7x-4}\]

Answer 5

Ok no this is A REALLY LONG QUESTION.

Answer 6

complete the square for the bottom,and double sub.

Answer 7

Long questions have never stopped you before :P

Answer 8

:( I shall call backup!

Answer 9

Partial fraction decomposition

Answer 10

Mr. Integral is here to integrate your knowledge and power.

Answer 11

Saw that right away

Answer 12

I can show what I got so far

Answer 13

Ah.

\[\large 2x^2+7x-4\]\[\large x^2+7x-8\]\[\large (x+8)(x-1)\]\[\large(x+4)(2x-1)\]

Answer 14

That breaks down into thaaaatt....

Answer 15

\[\large 5 \int\limits \frac{1}{(x+4)(2x-1)}= \frac{A}{x+4}+\frac{B}{2x-1}\]

Answer 16

Leave the 5 inside

Answer 17

(i'm just writing what comes to mind)

Answer 18

\[\frac{ 5 }{ (2x-1)(x+4) }=\frac{ A }{ 2x-1 }+\frac{ B }{ x+4 }\]
\[5=A(x+4)+B(2x-1)\]
\[=(A+2B)x+(4A-B)\]
\[A=\frac{ 10 }{ 9 }; B=\frac{ -5 }{ 9 }\]

Answer 19

skipped a few steps but I think you guys got it

Answer 20

where?

Answer 21

\[\frac{ 10 }{ 9 }\int\limits_{}^{}\frac{ 1 }{ 2x-1 }dx+\int\limits_{}^{}\frac{ 1 }{ x+4 }dx\]

Answer 22

Woops

Answer 23

\[\large \int\limits\limits \frac{5}{(x+4)(2x-1)}= \frac{A}{x+4}+\frac{B}{2x-1}\]\[\large 5=A(2x-1)+B(x+4)\]\[\large 5 = 2Ax-A+Bx+4B\]\[\large 5 = 2x(A+B)+(4B-A)\]

Answer 24

There's a constant of -5/9 before the 2nd integral