∫tan^(−1) (((a(√x)+b)/c))dx


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

$\int {t a n}^{- 1} (\frac{a \sqrt{x} + b}{c}) d x$
	Answered by Dwaipayan Shikari last updated on 23/Jul/20

	$\int \frac{2 c \sqrt{x}}{a} {t a n}^{- 1} u d u u = \frac{a \sqrt{x} + b}{c} \Rightarrow \frac{a}{2 c \sqrt{x}} = \frac{d u}{d x}$ $\frac{2 c}{a} \int \sqrt{x} {t a n}^{- 1} d u \sqrt{x} = \frac{c u - b}{a}$ $\frac{2 c}{a} \int (\frac{c u}{a} - \frac{b}{a}) {t a n}^{- 1} u d u$ $\frac{2 c^{2}}{a^{2}} \int {u t a n}^{- 1} u - \frac{2 b c}{a^{2}} \int {t a n}^{- 1} u d u$ $\int {u t a n}^{- 1} u = \frac{u^{2}}{2} {t a n}^{- 1} u - \frac{1}{2} \int \frac{u^{2}}{u^{2} + 1} = \frac{u^{2}}{2} {t a n}^{- 1} u - \frac{u}{2} - \frac{1}{2} {t a n}^{- 1} u$ $\int {t a n}^{- 1} u d u = {u t a n}^{- 1} u - \frac{1}{2} l o g (u^{2} + 1)$ $\frac{2 c^{2}}{a^{2}} (\frac{u^{2}}{2} {t a n}^{- 1} u - \frac{u}{2} - \frac{1}{2} {t a n}^{- 1} u) - \frac{2 b c}{a^{2}} ({u t a n}^{- 1} u - \frac{1}{2} l o g (u^{2} + 1)) + C$ $\frac{2 c^{2}}{a^{2}} (\frac{{(\frac{a \sqrt{x} + b}{c})}^{2}}{2} {t a n}^{- 1} (\frac{a \sqrt{x} + b}{c}) - \frac{a \sqrt{x} + b}{2 c} - \frac{1}{2} {t a n}^{- 1} (\frac{a \sqrt{x} + b}{c})) +$ $(- \frac{2 b c}{a^{2}} (\frac{a \sqrt{x} + b}{c} {t a n}^{- 1} (\frac{a \sqrt{x} + b}{c}) - \frac{1}{2} l o g ({(\frac{a \sqrt{x} + b}{c})}^{2} + 1)) + C$
	Commented by M±th+et+s last updated on 23/Jul/20

	$w e l l d o n e$
	Answered by mathmax by abdo last updated on 24/Jul/20
	$I =∫ arctan(((a(√x)+b)/c))dx we do the changement ((a(√x)+b)/c) =t ⇒ a(√x)+b =ct ⇒a(√x)=ct−b ⇒a^2 x =(ct−b)^2 ⇒x =(((ct−b)^2 )/a^2 ) ⇒ (dx/dt) =((2c(ct−b))/a^2 ) ⇒ I =((2c)/a^2 )∫ arctan(t) (ct−b)dt ⇒ (a^2 /(2c))×I =∫ (ct−b)arctan(t)dt =_(by parts) (((ct^2 )/2)−bt)arctan(t)−∫(((ct^2 )/2)−bt)(dt/(1+t^2 )) =(((ct^2 )/2)−bt)arctan(t)−(c/2) ∫ (t^2 /(1+t^2 ))dt +b ∫ ((tdt)/(1+t^2 )) we have ∫ (t^2 /(1+t^2 ))dt =∫ ((1+t^2 −1)/(1+t^2 ))dt =t−arctan(t)+c_0 ∫ ((tdt)/(1+t^2 )) =(1/2)ln(1+t^2 )+c_1 ⇒ (a^2 /(2c))I =(((ct^2 )/2)−bt)arctan(t)−(c/2){t−arctan(t)}+(b/2)ln(1+t^2 ) +C =((c/2)(((a(√x)+b)/c))^2 −b×((a(√x)+b)/c))arctan(((a(√x)+b)/c))−(c/2){((a(√x)+b)/c)−arctan(((a(√x)+b)/c)) +(b/2)ln(1+(((a(√x)+b)/c))^2 ) +C .$
	$I = \int \arctan (\frac{a \sqrt{x} + b}{c}) dx we do the changement \frac{a \sqrt{x} + b}{c} = t \Rightarrow$ $a \sqrt{x} + b = ct \Rightarrow a \sqrt{x} = ct - b \Rightarrow a^{2} x = {(ct - b)}^{2} \Rightarrow x = \frac{{(ct - b)}^{2}}{a^{2}} \Rightarrow$ $\frac{dx}{dt} = \frac{2 c (ct - b)}{a^{2}} \Rightarrow I = \frac{2 c}{a^{2}} \int \arctan (t) (ct - b) dt \Rightarrow$ $\frac{a^{2}}{2 c} \times I = \int (ct - b) \arctan (t) dt =_{by parts} (\frac{{ct}^{2}}{2} - bt) \arctan (t) - \int (\frac{{ct}^{2}}{2} - bt) \frac{dt}{1 + t^{2}}$ $= (\frac{{ct}^{2}}{2} - bt) \arctan (t) - \frac{c}{2} \int \frac{t^{2}}{1 + t^{2}} dt + b \int \frac{tdt}{1 + t^{2}} we have$ $\int \frac{t^{2}}{1 + t^{2}} dt = \int \frac{1 + t^{2} - 1}{1 + t^{2}} dt = t - \arctan (t) + c_{0}$ $\int \frac{tdt}{1 + t^{2}} = \frac{1}{2} \ln (1 + t^{2}) + c_{1} \Rightarrow$ $\frac{a^{2}}{2 c} I = (\frac{{ct}^{2}}{2} - bt) \arctan (t) - \frac{c}{2} {t - \arctan (t)} + \frac{b}{2} \ln (1 + t^{2}) + C$ $= (\frac{c}{2} {(\frac{a \sqrt{x} + b}{c})}^{2} - b \times \frac{a \sqrt{x} + b}{c}) \arctan (\frac{a \sqrt{x} + b}{c}) - \frac{c}{2} {\frac{a \sqrt{x} + b}{c} - \arctan (\frac{a \sqrt{x} + b}{c})$ $+ \frac{b}{2} \ln (1 + {(\frac{a \sqrt{x} + b}{c})}^{2}) + C .$
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