calculate-0-1-xlnx-x-1-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 33677 by math khazana by abdo last updated on 21/Apr/18 calculate∫01xlnxx−1dx. Commented by math khazana by abdo last updated on 30/Apr/18 I=∫01(x−1+1)x−1ln(x)dx=∫01ln(x)x+∫01lnxx−1dx∫01ln(x)dx=[xlnx−x]x→01=−1∫01ln(x)x−1dx=−∫01(∑n=0∞xn)ln(x)dx=−∑n=0∞∫01xnln(x)dxbypartsAn=∫01xnln(x)dx=[1n+1xn+1ln(x)]01−∫011n+1xndx=−1n+1∫01xndx=−1(n+1)2⇒∫01xln(x)x−1dx=−1+∑n=0∞1(n+1)2=∑n=1∞1n2−1∫01xln(x)x−1dx=π26−1. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: How-fast-is-the-height-of-a-balloon-changing-when-500m-away-at-an-angle-of-pi-4-rad-and-the-angle-is-increasing-by-0-2rad-min-Next Next post: Question-99218 Leave a Reply Cancel replyYour email address will not be published.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.
![I =∫_0 ^1 (((x−1+1))/(x−1))ln(x)dx = ∫_0 ^1 ln(x)x +∫_0 ^1 ((lnx)/(x−1))dx ∫_0 ^1 ln(x)dx=[xlnx −x]_(x→0) ^1 =−1 ∫_0 ^1 ((ln(x))/(x−1))dx=−∫_0 ^1 (Σ_(n=0) ^∞ x^n )ln(x)dx =−Σ_(n=0) ^∞ ∫_0 ^1 x^n ln(x)dx by parts A_n = ∫_0 ^1 x^n ln(x)dx=[(1/(n+1))x^(n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/(n+1)) x^n dx =−(1/(n+1)) ∫_0 ^1 x^n dx=−(1/((n+1)^2 )) ⇒ ∫_0 ^1 ((xln(x))/(x−1))dx= −1+Σ_(n=0) ^∞ (1/((n+1)^2 )) =Σ_(n=1) ^∞ (1/n^2 )−1 ∫_0 ^1 ((xln(x))/(x−1))dx = (π^2 /6) −1 .](https://www.tinkutara.com/question/Q34057.png)