calculate-0-1-ln-1-x-1-x-2-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 64065 by mathmax by abdo last updated on 12/Jul/19 calculate∫01ln(1+x)1+x2dx Commented by mathmax by abdo last updated on 13/Jul/19 letf(t)=∫01ln(1+tx)1+x2dxwitht>0wehavef′(t)=∫01x(1+tx)(1+x2)dxletdecomposeF(x)=x(tx+1)(x2+1)⇒F(x)=atx+1+bx+cx2+1a=limx→−1t(tx+1)F(x)=−1t(1t2+1)=−t2t(1+t2)=−tt2+1limx→+∞xF(x)=0=at+b⇒b=−at=1t2+1⇒F(x)=−t(t2+1)(tx+1)+xt2+1+cx2+1F(0)=0=tt2+1+c⇒c=−tt2+1⇒F(x)=t(t2+1)(tx+1)+1t2+1x−tx2+1⇒f′(t)=∫01F(x)dx=tt2+1∫01dxtx+1+1t2+1∫01x−tx2+1dx=tt2+11t[ln(tx+1)]01+12(t2+1)∫012xx2+1dx−tt2+1∫01dxx2+1=ln(t+1)t2+1+12(t2+1)ln(2)−tt2+1π4⇒f(t)=∫0tln(1+u)u2+1du+ln(2)2∫0tdu1+u2−π8∫0t2u1+u2du+c=∫0tln(1+u)1+u2+ln(2)2arctan(t)−π8ln(1+t2)+c=∫01ln(1+tx)1+x2dx…becontinued… Commented by mathmax by abdo last updated on 13/Jul/19 f(1)=∫01ln(1+x)1+x2dx=∫01ln(1+u)1+u2du+π8ln(2)−π8ln(2)+c⇒c=0and∫01ln(1+tx)1+x2dx=∫0tln(1+x)1+x2dx+ln(2)2arctan(t)−π8ln(1+t2)…becontinued… Commented by mathmax by abdo last updated on 14/Jul/19 lettryanotherwaychangementx=tanθgiveI=∫01ln(1+x)1+x2dx=∫0π4ln(1+tanθ)1+tan2θ(1+tan2θ)dθ=∫0π4ln(1+tanθ)dθchangθ=π4−u⇒∫0π4ln(1+tanθ)dθ=∫π40ln(1+1−tanu1+tanu)(−du)=∫0π4ln(21+tanu)du=π4ln(2)−∫0π4ln(1+tanu)du⇒I=π4ln(2)−I⇒2I=π4ln(2)⇒I=π8ln(2) Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 2x-3-2-25-x-3-2-2-Next Next post: let-and-the-roots-of-x-3-2x-1-0-find-the-value-of-A-2-2-2-and-B-3-3-3- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.