Integrate the function  \frac{2x}{1 + {x}^{2} }

Question

Integrate the function
 \frac{2x}{1 +  {x}^{2} }

in progress 0
Lyla 2 months 2021-09-25T06:39:46+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-09-25T06:41:26+00:00

     \rm \implies \displaystyle \int \rm \dfrac{2x}{1 + {x}^{2} } \: dx \\  \\  \\ \rm \implies \displaystyle \int \rm \dfrac{2x \:dx }{1 + {x}^{2} } \:  \\  \\  \\  \rm \: let \: \:  1 + {x}^{2} = t \\  \rm0 + 2x \: dx = dt\\  \rm 2x \: dx = dt \\  \\  \\ \rm \implies \displaystyle \int \rm \dfrac{dt}{t } \\  \\  \\ \rm \implies \displaystyle \rm log \: t + c\\  \\ \\   \rm \: now \: put \: t = 1 +  {x}^{2}   \\  \\ \\ \rm \implies \displaystyle \rm log \: (1 +  {x}^{2})  + c

    Additional Information :

    \boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}

    0
    2021-09-25T06:41:40+00:00

    \large\underline\blue{\bold{Given \:  Question :-  }}

    \bf \:Evaluate :  \: \bf\int\limits \dfrac{2xdx}{1+x^2}

    ─━─━─━─━─━─━─━─━─━─━─━─━─

    \huge{AηsωeR} ✍

    ─━─━─━─━─━─━─━─━─━─━─━─━─

    \bf \:  ⟼ Let \:  I =\bf\int\limits\dfrac{2x}{1 +  {x}^{2} } dx

    \bf \:Put \:  {1 + x}^{2}  = t

    ☆ Differentiate w. r. t. x, we get

    \bf \:\dfrac{d}{dx}  (1 +  {x}^{2}) = \dfrac{d}{dx} t

    \bf\implies \:2x = \dfrac{dt}{dx}

    \bf\implies \:2xdx = dt

    \bf \:  ⟼ so

    \bf \: I =\bf\int\limits\dfrac{1}{t} dt

    \bf\implies \: I =logt \:  + c

    \bf\implies \: I = log(1 +  {x}^{2} )  + c

    \large{\boxed{\boxed{\bf{\bf\int\limits \dfrac{2xdx}{1+x^2} = log(1 +  {x}^{2} )  + c}}}}

    ─━─━─━─━─━─━─━─━─━─━─━─━─

    \huge {AlteR \: method} ✍

    ☆ We know,

    \bf\int\limits\dfrac{f'(x)}{f(x)}  =  log(f(x))  + c

    ─━─━─━─━─━─━─━─━─━─━─━─━─

    \bf \:Let \:  I =\bf\int\limits \dfrac{2xdx}{1+x^2}

    \bf\implies  I =\:\bf\int\limits \dfrac{\dfrac{d}{dx} ( {1 + x}^{2} )}{1+x^2}dx

    \bf\implies \: I =log(1 +  {x}^{2} )  + c

    \large{\boxed{\boxed{\bf{\bf\int\limits \dfrac{2xdx}{1+x^2} = log(1 +  {x}^{2} )  + c}}}}

    ─━─━─━─━─━─━─━─━─━─━─━─━─

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )