Flag Integral Calculus> Integration of log(1+ax) /1+x² , limits f...
question mark

Integration of log(1+ax) /1+x² , limits from 0 to a

Pranav , 4 Years ago
Grade 12
anser 1 Answers
Manika gupta

Last Activity: 4 Years ago

Dear student
See you solution below

(d/da) ∫(x = 0 to a) [log(1+ax)/(1+x^2)] dx

= ∫(x = 0 to a) (∂/∂a) [log(1+ax)/(1+x^2)] dx + [log(1+a^2)/(1+a^2)] * (d/da)(a) - [log(1+0)/(1+0)] * (d/da)0

= log(1+a^2)/(1+a^2) + ∫(x = 0 to a) x dx/[(1+ax)(1+x^2)]

= log(1+a^2)/(1+a^2) + (1/(a^2+1)) ∫(x = 0 to a) [-a/(1+ax) + (x+a)/(1+x^2)] dx

= log(1+a^2)/(1+a^2) + (1/(a^2 + 1)) [-log(1+ax) + ((1/2) log(1+x^2) + a arctan x)] {x = 0 to a}

= (1/2) log(1+a^2)/(1+a^2) + a arctan a/(1 + a^2).

-----------

Since we have

(d/da) ∫(x = 0 to a) [log(1+ax)/(1+x^2)] dx = (1/2) log(1+a^2)/(1+a^2) + a arctan a/(1+a^2)

integrating both sides with respect to a yields

∫(x = 0 to a) [log(1+ax)/(1+x^2)] dx

= ∫ [(1/2) log(1+a^2)/(1+a^2)] da + ∫ [a arctan a/(1+a^2)] da

= {(1/2) log(1+a^2) arctan a - ∫ [a arctan a/(1+a^2)] da} + ∫ [a arctan a/(1+a^2)] da

= (1/2) log(1+a^2) arctan a + C.

----

To find C, note that letting a = 0 yields ∫(x = 0 to 0) [log(1+0)/(1+x^2)] dx = 0.

Hence, 0 = (1/2) log(1+0) arctan 0 + C ==> C = 0.

Therefore, ∫(x = 0 to a) [log(1+ax)/(1+x^2)] dx = (1/2) log(1+a^2) arctan a.

Askiitians expert

Provide a better Answer & Earn Cool Goodies

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free