IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

jayanthaliasalonso

x³/(a+bx)²

dr.dane

put a+bx = t
differentiating
b dx = dt
dx = dt/b
putting in given equation
1?(b²)²?(t-a)³?t²

?( t³-a³-3t²a+3a²t)÷t²(b²)² dt

[(t²\2) + (a³/t) ? 3at + 3a²logt]÷(b²)²

on putting value of t in this equation i.e. t=a+bx u get the result

plss rate me

priyesh

put a + bx = t

dx = dt/b

also x = (t -a)/b

therefore integral becomes

1/b [(t-a)/b]³] / t²]dt

= 1/b⁴ (t-a)³/t² dt

= 1/b⁴ [t³ - a³ - 3at² + 3 a²t]/t²

= 1/b⁴ [t -a³1/t² - 3a + 3a² 1/t] dt

integrating term by term

= 1/b⁴ [

t -a³

1/t² - 3a

1 + 3a²

1/t] dt

= 1/b⁴ [ t²/2 + a³/t - 3at +3a²lnt]

now replace t by a + bx to get the answer

iitkgp_bipin

Well done guys.

Substitute ax+b = t, you will have different terms of t with different powers which can easily be integrated. At last back substitute t to get the result.

jatinroxx

Its an improper integral...

Convert it into a proper one, its a piece of cake then.......

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