The correct option is C e^x.1/1+x^2. ( 1 x/1+x^2 )^2=1+x^2 2x/ ( 1+x^2 )^2 1/1+x^2+ 2x/ ( 1+x^2 )^2=f ( x )+f' ( x ) ∴ I=∫ e^x [ f ( x )+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Higher Order Derivatives

∫ ex 1-x/1+x2...

Question

$\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x .$

e^{x} . \frac{- 1}{1 + x^{2}} .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{x} . \frac{1}{1 - x^{2}} .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{x} . \frac{1}{1 + x^{2}} .

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{2} e^{x} . \frac{1}{1 + x^{2}} .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $e^{x} . \frac{1}{1 + x^{2}} .$
${(\frac{1 - x}{1 + x^{2}})}^{2} = {\frac{1 + x^{2} - 2 x}{(1 + x^{2})}}^{2}$
$\frac{1}{1 + x^{2}} + \frac{- 2 x}{{(1 + x^{2})}^{2}} = f (x) + f^{'} (x)$
$∴ I = \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} . f (x) = e^{x} . \frac{1}{1 + x^{2}} .$

Suggest Corrections

Similar questions

$\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2}$ dx

$\int \frac{x e^{x}}{(1 + x)^{2}} d x =$

[\frac{1}{x^{2}} + \frac{{(2013)}^{x}}{e^{x} - 1} - \frac{1}{e^{x} - 1}]

x \to 0

[\frac{1}{x^{2}} + \frac{(2013)^{x}}{e^{x} - 1} - \frac{1}{e^{x} - 1}]

x \to 0

\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x is equal to (a) \frac{e^{x}}{1 + x} + C (b) - \frac{e^{x}}{1 + x^{2}} + C (c) \frac{e^{x}}{(1 + x}

Join BYJU'S Learning Program