Question

Pls answer asap! I'll mark u the brainliest!​

Answer 1

Given,

$\small\text{ \displaystyle\longrightarrow\int\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}\,dx=k\log\left|\tan^{-1}\left(\dfrac{x^2+1}{x}\right)\right|+c }$

Differentiating both sides wrt x,

$\small\text{ \longrightarrow\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=\dfrac{d}{dx}\left[k\log\left|\tan^{-1}\left(\dfrac{x^2+1}{x}\right)\right|+c\right] }$

We have,

• $\small\dfrac{d}{dx}(c\cdot f(x))=c\cdot\dfrac{d}{dx}(f(x))$

• $\small\text{ \dfrac{d}{dx}\big[\log|f(x)|\big]=\dfrac{\dfrac{d}{dx}(f(x))}{f(x)}, }$

• $\small\dfrac{d}{dx}(c)=0$

where c is a constant. Then,

$\small\text{ \longrightarrow\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=k\cdot\dfrac{\dfrac{d}{dx}\left[\tan^{-1}\left(\dfrac{x^2+1}{x}\right)\right]}{\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}+0 }$

$\small\text{ \longrightarrow\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=k\cdot\dfrac{\dfrac{d}{dx}\left[\tan^{-1}\left(x+\dfrac{1}{x}\right)\right]}{\tan^{-1}\left(\dfrac{x^2+1}{x}\right)} }$

By chain rule,

• $\small\dfrac{d}{dx}(f(g(x)))=\dfrac{d}{d(g(x))}(f(g(x)))\cdot\dfrac{d}{dx}(g(x))$

Then,

$\small\text{ \longrightarrow\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=k\cdot\dfrac{\dfrac{1}{1+\left(x+\dfrac{1}{x}\right)^2}\cdot\left(1-\dfrac{1}{x^2}\right)}{\tan^{-1}\left(\dfrac{x^2+1}{x}\right)} }$

$\small\text{ \longrightarrow\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=k\cdot\dfrac{\dfrac{1}{x^2+\dfrac{1}{x^2}+3}\cdot\left(1-\dfrac{1}{x^2}\right)}{\tan^{-1}\left(\dfrac{x^2+1}{x}\right)} }$

$\small\text{ \longrightarrow\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=k\cdot\dfrac{1-\dfrac{1}{x^2}}{\left(x^2+\dfrac{1}{x^2}+3\right)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)} }$

Multiplying both numerator and denominator of the fraction in RHS by $\smallx^2,$

$\small\text{ \longrightarrow\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=k\cdot\dfrac{x^2\left(1-\dfrac{1}{x^2}\right)}{x^2\left(x^2+\dfrac{1}{x^2}+3\right)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)} }$

$\small\text{ \longrightarrow\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=k\cdot\dfrac{x^2-1}{\left(x^4+3x^2+1\right)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)} }$

$\small\text{ \longrightarrow(k-1)\cdot\dfrac{x^2-1}{\left(x^4+3x^2+1\right)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}=0 }$

Since $\small\text{ \dfrac{x^2-1}{\left(x^4+3x^2+1\right)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}\neq0 }$ for all x, we get,

$\small\text{ \longrightarrow\underline{\underline{k=1}} }$

Hence 1 is the answer.

Answer 2

Given Question :-

If

$\small\text{ \displaystyle\int\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}\,dx=k\log\left|\tan^{-1}\left(\dfrac{x^2+1}{x}\right)\right|+c }$

then the value of k is equals to

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\displaystyle\int\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}\,dx$

$\rm \:  =  \: k\log\left|\tan^{-1}\left(\dfrac{x^2+1}{x}\right)\right|+c$

Let assume that,

$\rm :\longmapsto\:I \: = \: \displaystyle\int\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\dfrac{x^2+1}{x}\right)}\,dx$

To evaluate this, we use method of Substitution.

Let we substitute,

$\red{\rm :\longmapsto\: {tan}^{ - 1}\dfrac{ {x}^{2} + 1}{x} = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\: \dfrac{d}{dx} {tan}^{ - 1}\dfrac{ {x}^{2} + 1}{x} = \dfrac{d}{dx} y}$

$\red{\rm :\longmapsto\:\dfrac{1}{1 + {\bigg[\dfrac{ {x}^{2} + 1}{x} \bigg]}^{2} }\dfrac{d}{dx}\dfrac{ {x}^{2} + 1}{x} = \dfrac{dy}{dx} }$

$\red{\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {x}^{2}+ {x}^{4} + 1 + {2x}^{2} }\dfrac{d}{dx}\bigg[x + \dfrac{1}{x} \bigg] = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:\dfrac{ {x}^{2} }{{x}^{4} + 1 + {3x}^{2} }\bigg[1 - \dfrac{1}{ {x}^{2} } \bigg] = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:\dfrac{ {x}^{2} }{{x}^{4} + 1 + {3x}^{2} }\bigg[\dfrac{ {x}^{2} - 1}{ {x}^{2} } \bigg] = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:\dfrac{ {x}^{2} - 1 }{{x}^{4} + 1 + {3x}^{2} } = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:\dfrac{ {x}^{2} - 1 }{{x}^{4} + {3x}^{2} + 1 } \: dx = dy}$

So, on substituting the values, we get

$\rm :\longmapsto\:I = \displaystyle\int \: \frac{dy}{y}$

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

$\bf\implies \:I = log\bigg | {tan}^{ - 1}\bigg( \dfrac{ {x}^{2} + 1}{x} \bigg)\bigg| + c$

But it is given that,

$\bf :\longmapsto\:I = k\log\left|\tan^{-1}\left(\dfrac{x^2+1}{x}\right)\right|+c$

So, on comparing, we get

$\red{\bf\implies \: \boxed{ \bf{ \:k \: = \: 1 \: \: }}}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$