Question

1.∫(2u+10)/(3u+13) du
2.∫(u+5)/(5u-2)^2 du

Answer 1

\[I = \int\limits{\frac{u+5}{(5u-2)^2}}du = \int\limits{\frac{A}{(5u-2)^2 } + \frac{B}{(5u-2)}}du \]

A + B(5u-2) = u+5

let u = 2/5

A = 27/5

Answer 2

let u = 0

27/5 -2B = 5

B = 1/5

Answer 3

For the first one do long division first

Answer 4

eigen : A + B(5u-2) = u+5 why not A(5u-2)^2 + B(5u-2) =u+5
lalaly : complete answer pls I'm too slow --"

Answer 5

Do u know how to do long divisio?

Answer 6

if
\[\frac{u+5}{(5u-2)^2} = \frac{A}{(5u-2)^2} + \frac{B}{5u-2}\]

multiply both sides by the LHS denominator to get the equation i had

Answer 7

lana tell him the easier way out
split into
2u/(10u+13) + 10/(10u+13)

Answer 8

Ok but still how would u integrate the first one @wasiqss

Answer 9

2u/(10u+13)?

Answer 10

yes lana imma tell you

Answer 11

Either way u should do long division @wasiqss

Answer 12

hmm partial fracction, but yehh it will lead us to same route

Answer 13

yayyyyyyy i found the easier method

Answer 14

multiply divide by 5

Answer 15

take 1/5 as common

Answer 16

lalaly's solution (i think is fairly simple anyway)


wasiqss:
2u/(10u+13)
will integrate by parts, not really worth it though