Question Number 101783 by I want to learn more last updated on 04/Jul/20

∫_( 0) ^( 1)   ((ln(x^2  + 1))/(x  +  1))

Answered by mathmax by abdo last updated on 04/Jul/20

let f(a) =∫_0 ^1  ((ln(1+ax^2 ))/(x+1))  with a>0  f^′ (a) =∫_0 ^1  (x^2 /((1+ax^2 )(x+1))) =(1/a)∫_0 ^1  ((1+ax^2 −1)/((1+ax^2 )(x+1)))dx  =(1/a)∫_0 ^1  (dx/(x+1)) −(1/a)∫_0 ^1  (dx/((1+ax^2 )(x+1)))  ((√a)x =u)  =(1/a)[ln∣x+1∣]_0 ^1  −(1/(a(√a))) ∫_0 ^(√a)   (du/((1+u^2 )((u/(√a)) +1)))  =(1/a)ln(2)−(1/a) ∫_0 ^(√a)   (du/((u^2  +1)(u+(√a)))) let decompose F(u) =(1/((u^2  +1)(u+(√a))))  F(u) =((αu+β)/(u^2  +1)) +(c/(u+(√a)))  c =(1/(a+1))   ,lim_(u→+∞) uF(u) =0 =α +c ⇒α =−(1/(a+1))  ⇒F(o) =(1/(√a)) =β +(c/(√a)) ⇒β =(1/(√a))−(c/(√a)) =(1/(√a))(1−(1/(a+1))) =(a/((√a)(a+1))) =((√a)/(a+1))  ⇒F(u) =((−(u/(a+1))+(1/(a+1)))/(u^2  +1)) +(1/((a+1)(u+(√a)))) ⇒  ∫_0 ^(√a)  F(u)du =−(1/(a+1)) ∫_0 ^(√a)  ((u−1)/(u^2  +1))du +(1/(a+1))∫_0 ^(√a)   (du/(u+(√a)))  =−(1/(2(a+1)))[ln(1+u^2 )]_0 ^(√a)  +(1/(a+1))[arctanu]_0 ^(√a)  −(1/(a+1))[ln(u+(√a))]_0 ^(√a)   =−(1/(2(a+1)))ln(1+a)+(1/(a+1))arctan((√a))−(1/(a+1)){ln(2(√a))−ln((√a))} ⇒  f(a)=((ln(2))/a) −(1/2) ∫ ((ln(1+a))/(a(a+1)))da−∫ ((arctan((√a)))/(a(a+1)))da −ln(2)∫  (da/(a(a+1))) +c  be continued....

Commented byI want to learn more last updated on 05/Jul/20

Please help me with just one thing..  What type of integration is this, what topic exactly  can i learn integrals like this.

Commented bymathmax by abdo last updated on 05/Jul/20

its a paremetric integration...

Commented bymathmax by abdo last updated on 05/Jul/20

thank you sir for giving this explanation...

Answered by Dwaipayan Shikari last updated on 05/Jul/20

If the question is ∫_0 ^1 ((log(x+1))/(x^2 +1))  Then ∫_0 ^1 ((log(tanθ+1))/(sec^2 θ))sec^2 θdθ    {  suppose x=tanθ  ∫_0 ^(π/4) log(sinθ+cosθ)−log(cosθ)dθ  ∫_0 ^(π/4) log((√2)(cos((π/4)−θ)))−log(cosθ)  ∫_0 ^(π/4) log(√2)+log(cos((π/4)−θ))−log(cosθ)=I    {∫_0 ^(π/4) f(θ)=∫_0 ^(π/4) f((π/4)−θ)  ∫_0 ^(π/4) log(√2)+log(cos(θ))−log(cos((π/4)−θ)=I   {adding }  2I=2∫_0 ^(π/4) log(√2)dθ  I=(1/2)log2[θ]_0 ^(π/4)   I=(π/8)log2

Commented bymathmax by abdo last updated on 05/Jul/20

this is not the question...!

Commented byDwaipayan Shikari last updated on 05/Jul/20

I have edited it.But sir you are right,this is not the question.  I am trying .

Commented by1549442205 last updated on 05/Jul/20

this is not the question...!  Thank you both sir a lot for giving the good suggest!