tan-1-a-x-b-c-dx-

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 = \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 .