Question

Evaluate the following integrals..
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Please answer fastt
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Who will give correct answer will be marked branialist​

Answer 1

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

Evaluate the following integral

$\displaystyle\int\sf \:\bigg[\dfrac{x + 1}{x} - \dfrac{2}{ \sqrt{1 - {x}^{2} } } - {2}^{2x} \bigg] \: dx$

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

The given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \:\bigg[\dfrac{x + 1}{x} - \dfrac{2}{ \sqrt{1 - {x}^{2} } } - {2}^{2x} \bigg] \: dx$

can be rewritten as

$\sf \: = \: \displaystyle\int\sf \bigg[\dfrac{x}{x} + \dfrac{1}{x} - \dfrac{2}{ \sqrt{1 - {x}^{2} } } + {4}^{x}\bigg] \: dx$

$\sf \: = \: \displaystyle\int\sf \bigg[1 + \dfrac{1}{x} - \dfrac{2}{ \sqrt{1 - {x}^{2} } } + {4}^{x}\bigg] \: dx$

$\sf \: = \: \displaystyle\int\sf 1dx + \displaystyle\int\sf \frac{1}{x}dx - 2\displaystyle\int\sf \frac{1}{ \sqrt{1 - {x}^{2} } }dx - \displaystyle\int\sf {4}^{x} dx$

$\sf \: = \: x + logx - 2 {sin}^{ - 1}x - \dfrac{ {4}^{x} }{log4} + c$

$\sf \: = \: x + logx - 2 {sin}^{ - 1}x - \dfrac{ {4}^{x} }{log {2}^{2} } + c$

$\red{\bf \: = \: x + logx - 2 {sin}^{ - 1}x - \dfrac{ {4}^{x} }{2log {2}} + c}$

Additional Information :-

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

Answer 2

Answer:

\large\underline{\bold{Given \:Question - }}GivenQuestion−

Evaluate the following integral

\displaystyle\int\sf \:\bigg[\dfrac{x + 1}{x} - \dfrac{2}{ \sqrt{1 - {x}^{2} } } - {2}^{2x} \bigg] \: dx∫[xx+1−1−x22−22x]dx

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

The given integral is

\rm :\longmapsto\:\displaystyle\int\sf \:\bigg[\dfrac{x + 1}{x} - \dfrac{2}{ \sqrt{1 - {x}^{2} } } - {2}^{2x} \bigg] \: dx:⟼∫[xx+1−1−x22−22x]dx

can be rewritten as

\sf \: = \: \displaystyle\int\sf \bigg[\dfrac{x}{x} + \dfrac{1}{x} - \dfrac{2}{ \sqrt{1 - {x}^{2} } } + {4}^{x}\bigg] \: dx=∫[xx+x1−1−x22+4x]dx

\sf \: = \: \displaystyle\int\sf \bigg[1 + \dfrac{1}{x} - \dfrac{2}{ \sqrt{1 - {x}^{2} } } + {4}^{x}\bigg] \: dx=∫[1+x1−1−x22+4x]dx

\sf \: = \: \displaystyle\int\sf 1dx + \displaystyle\int\sf \frac{1}{x}dx - 2\displaystyle\int\sf \frac{1}{ \sqrt{1 - {x}^{2} } }dx - \displaystyle\int\sf {4}^{x} dx=∫1dx+∫x1dx−2∫1−x21dx−∫4xdx

\sf \: = \: x + logx - 2 {sin}^{ - 1}x - \dfrac{ {4}^{x} }{log4} + c=x+logx−2sin−1x−log44x+c

\sf \: = \: x + logx - 2 {sin}^{ - 1}x - \dfrac{ {4}^{x} }{log {2}^{2} } + c=x+logx−2sin−1x−log224x+c

\red{\bf \: = \: x + logx - 2 {sin}^{ - 1}x - \dfrac{ {4}^{x} }{2log {2}} + c}=x+logx−2sin−1x−2log24x+c

Additional Information :-

\begin{gathered}\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}\end{gathered}f(x)ksinxcosxsec2xcosec2xsecxtanxcosecxcotxtanxx1ex∫f(x)dxkx+c−cosx+csinx+ctanx+c−cotx+csecx+c−cosecx+clogsecx+clogx+cex+c