Question Number 104104 by bemath last updated on 19/Jul/20

∫ ((xtan^(−1) (x))/(√(1+x^2 ))) dx ?

Answered by OlafThorendsen last updated on 19/Jul/20

By parts :  (√(1+x^2 ))arctanx−∫(√(1+x^2 ))(dx/(1+x^2 ))  (√(1+x^2 ))arctanx−∫(dx/(√(1+x^2 )))  (√(1+x^2 ))arctanx−argshx+C

Answered by Dwaipayan Shikari last updated on 19/Jul/20

∫((θtanθ)/(secθ)) sec^2 θdθ                          {x=tanθ  ⇒1=sec^2 θ(dθ/dx)  ∫θsecθtanθ  θ∫secθtanθ−∫∫secθtanθ  θsecθ−log(secθ+tanθ)+C  tan^(−1) x sec(tan^(−1) x)−log(sec(tan^(−1) x)+x)+C      Another way      tan^(−1) x.(1/2)∫((2x)/(√(x^2 +1)))−∫(1/(x^2 +1)).(1/2)∫((2x)/(√(x^2 +1)))  tan^(−1) x(√(x^2 +1))−∫(dx/(√(x^2 +1)))=tan^(−1) x (√(x^2 +1))−log(x+(√(x^2 +1)))+C

Answered by bobhans last updated on 19/Jul/20

∫ tan^(−1) (x) d((√(1+x^2 ))) =   (√(1+x^2 )) tan^(−1) (x)−∫ ((√(1+x^2 ))/(1+x^2 )) dx =  (√(1+x^2 )) tan^(−1) (x)−∫ (dx/(√(1+x^2 )))  let I_1 = ∫ (dx/(√(1+x^2 )))  [ x = tan p ]  I_1 = ∫ ((sec ^2 p)/(sec p)) dp = ∫ sec p dp  I_1 = ln ∣sec p + tan p ∣ + c   ∴ I = (√(1+x^2 )) tan^(−1) (x)−ln ∣(√(1+x^2 )) +x∣ + c

Answered by mathmax by abdo last updated on 19/Jul/20

I =∫  ((xarctanx)/(√(1+x^2 )))dx   by parts u^′  =(x/(√(1+x^2 ))) and v =arctanx ⇒  I =(√(1+x^2 )) arctanx −∫ (√(1+x^2 ))×(dx/(1+x^2 )) =arctanx(√(1+x^2 )) −∫ (dx/(√(1+x^2 )))  =arctan(x)(√(1+x^2 )) −ln(x+(√(1+x^2 ))) +C