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

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

Grade:12

1 Answers

Manika gupta
askIITians Faculty 43 Points
3 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).

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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.

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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

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