Question Number 104718 by  M±th+et+s last updated on 23/Jul/20

∫tan^(−1) (((a(√x)+b)/c))dx

Answered by Dwaipayan Shikari last updated on 23/Jul/20

∫((2c(√x))/a)tan^(−1) udu                                          u=((a(√x)+b)/c)⇒(a/(2c(√x)))=(du/dx)  ((2c)/a)∫(√x)tan^(−1) du                                        (√x)=((cu−b)/a)  ((2c)/a)∫(((cu)/a)−(b/a))tan^(−1) udu  ((2c^2 )/a^2 )∫utan^(−1) u−((2bc)/a^2 )∫tan^(−1) udu  ∫utan^(−1) u=(u^2 /2)tan^(−1) u−(1/2)∫(u^2 /(u^2 +1))=(u^2 /2)tan^(−1) u−(u/2)−(1/2)tan^(−1) u  ∫tan^(−1) udu=utan^(−1) u−(1/2)log(u^2 +1)  ((2c^2 )/a^2 )((u^2 /2)tan^(−1) u−(u/2)−(1/2)tan^(−1) u)−((2bc)/a^2 )(utan^(−1) u−(1/2)log(u^2 +1))+C  ((2c^2 )/a^2 )((((((a(√x)+b)/c))^2 )/2)tan^(−1) (((a(√x)+b)/c))−((a(√x)+b)/(2c))−(1/2)tan^(−1) (((a(√x)+b)/c)))+  (−((2bc)/a^2 )(((a(√x)+b)/c)tan^(−1) (((a(√x)+b)/c))−(1/2)log((((a(√x)+b)/c))^2 +1))+C

Commented by M±th+et+s last updated on 23/Jul/20

well done

Answered by mathmax by abdo last updated on 24/Jul/20

I =∫ arctan(((a(√x)+b)/c))dx  we do the changement ((a(√x)+b)/c) =t ⇒  a(√x)+b =ct ⇒a(√x)=ct−b ⇒a^2  x =(ct−b)^2  ⇒x =(((ct−b)^2 )/a^2 ) ⇒  (dx/dt) =((2c(ct−b))/a^2 ) ⇒ I =((2c)/a^2 )∫ arctan(t) (ct−b)dt ⇒  (a^2 /(2c))×I =∫ (ct−b)arctan(t)dt =_(by parts)   (((ct^2 )/2)−bt)arctan(t)−∫(((ct^2 )/2)−bt)(dt/(1+t^2 ))  =(((ct^2 )/2)−bt)arctan(t)−(c/2) ∫ (t^2 /(1+t^2 ))dt +b ∫ ((tdt)/(1+t^2 ))  we have  ∫ (t^2 /(1+t^2 ))dt =∫ ((1+t^2 −1)/(1+t^2 ))dt =t−arctan(t)+c_0   ∫  ((tdt)/(1+t^2 )) =(1/2)ln(1+t^2  )+c_1  ⇒  (a^2 /(2c))I =(((ct^2 )/2)−bt)arctan(t)−(c/2){t−arctan(t)}+(b/2)ln(1+t^2 ) +C  =((c/2)(((a(√x)+b)/c))^2 −b×((a(√x)+b)/c))arctan(((a(√x)+b)/c))−(c/2){((a(√x)+b)/c)−arctan(((a(√x)+b)/c))  +(b/2)ln(1+(((a(√x)+b)/c))^2 ) +C .