Question

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

Answer 1

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

Additional Information :

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

Answer 2

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

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

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$\huge{AηsωeR} ✍$

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

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

☆ Differentiate w. r. t. x, we get

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

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

$\bf\implies \:2xdx = dt$

$\bf \:  ⟼ so$

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

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

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

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

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$\huge {AlteR \: method} ✍$

☆ We know,

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

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

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

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

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

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