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Model Paper 7 — Applied Maths STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceApplied MathsModel Paper 71 Mark
MCQ
$\int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x$ is equal to
  • A
    $-\frac{e^x}{1+x^2}+C$
  • B
    $-\frac{e^x}{\left(1+x^2\right)^2}+C$
  • ✓
    $\frac{e^x}{1+x^2}+C$
  • D
    $\frac{e^x}{\left(1+x^2\right)^2}+C$
✓

Answer

Correct option: C.
$\frac{e^x}{1+x^2}+C$
(c) $\frac{e^x}{1+x^2}+C$
$\begin{array}{l}\text { Explanation: Given } \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x \\ \Rightarrow \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x=\int e^x\left(\frac{1+x^2-2 x}{\left(1+x^2\right)^2}\right) d x \\ \Rightarrow \int e^x\left(\frac{1+x^2-2 x}{\left(1+x^2\right)^2}\right) d x=\int e^x\left\{\left(\frac{1+x^2}{\left(1+x^2\right)^2}\right)+\left(\frac{-2 x}{\left(1+x^2\right)^2}\right)\right\} d x \\ =\int e^x\left\{\left(\frac{1}{\left(1+x^2\right)}\right)+\left(\frac{-2 x}{\left(1+x^2\right)^2}\right)\right\} d x\end{array}$
Now using the property: $\int e ^{ x }\left( f ( x )+ f ^{\prime}( x )\right) dx = e ^{ x } f ( x )$
Now in $\int e^x\left\{\left(\frac{1}{\left(1+x^2\right)}\right)+\left(\frac{-2 x}{\left(1+x^2\right)^2}\right)\right\} d x$
$\begin{array}{l}\Rightarrow f ( x )=\frac{1}{\left(1+x^2\right)} \\ \Rightarrow f ^{\prime}( x )=\frac{-2 x}{\left(1+x^2\right)^2} \\ \Rightarrow \int e ^{ x }\left\{\left(\frac{1}{\left(1+ x ^2\right)}\right)+\left(\frac{-2 x }{\left(1+ x ^2\right)^2}\right)\right\} dx =\frac{ e ^{ x }}{1+ x ^2}+ C \\ \Rightarrow \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x=\frac{e^x}{1+x^2}+C\end{array}$

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