JEE Main 2013 22 April Online
Exam JEE Main
Year 2013
Date 22 April
Mode Online
Subject Mathematics
Chapter Indefinite Integration JEE Main 2013 · 22 April · Online · Mathematics · Indefinite Integration
If ∫ (x^2-x+1)/(x^2+1) e^ ^ -1 x d x=A(x) e^ ^ -1 x +C , then A(x)...
If
∫ x 2 − x + 1 x 2 + 1 e cot − 1 x d x = A ( x ) e cot − 1 x + C \int \frac{x^2-x+1}{x^2+1} e^{\cot ^{-1} x} d x=A(x) e^{\cot ^{-1} x}+C ∫ x 2 + 1 x 2 − x + 1 e c o t − 1 x d x = A ( x ) e c o t − 1 x + C , then
A ( x ) A(x) A ( x ) is equal to :
Options A − x -x − x B x x x C 1 − x \sqrt{1-x} 1 − x D 1 + x \sqrt{1+x} 1 + x
Step-by-step solution
Let
I = ∫ x 2 − x + 1 x 2 + 1 ⋅ e cot − 1 x d x \mathrm{I}=\int \frac{x^2-x+1}{x^2+1} \cdot e^{\cot ^{-1} x} d x I = ∫ x 2 + 1 x 2 − x + 1 ⋅ e c o t − 1 x d x
Put
x = cot t ⇒ − cosec 2 t d t = d x x=\cot t \Rightarrow-\operatorname{cosec}^2 t d t=d x x = cot t ⇒ − cosec 2 t d t = d x
Now,
1 + cot 2 t = cosec 2 t ∴ 1+\cot ^2 t=\operatorname{cosec}^2 t\therefore 1 + cot 2 t = cosec 2 t ∴ \mathrm{I}
= ∫ e t ( cot 2 t − cot t + 1 ) ( 1 + cot 2 t ) ( − cosec 2 t ) d t =\int \frac{e^t\left(\cot ^2 t-\cot t+1\right)}{\left(1+\cot ^2 t\right)}\left(-\operatorname{cosec}^2 t\right) d t = ∫ ( 1 + c o t 2 t ) e t ( c o t 2 t − c o t t + 1 ) ( − cosec 2 t ) d t = − \inte t ( cosec 2 t − cot t ) d t = \inte t ( \cott − cosec 2 t ) d t = e t cot t + C = e cot − 1 x ( x ) + C ≡ A ( x ) \cdote cot − 1 x + C ⇒ A ( x ) = x \begin{aligned}& =-\inte^t\left(\operatorname{cosec}{ }^2 t-\cot t\right) d t \\
& =\inte^t\left(\cott-\operatorname{cosec}^2 t\right) d t \\
& =e^t \cot t+\mathrm{C}\\
& =e^{\cot ^{-1} x}(x)+\mathrm{C} \equiv \mathrm{A}(x) \cdote^{\cot ^{-1} x}+\mathrm{C}\\
\Rightarrow \mathrm{A}(x) & =x
\end{aligned} ⇒ A ( x ) = − \inte t ( cosec 2 t − cot t ) d t = \inte t ( \cott − cosec 2 t ) d t = e t cot t + C = e c o t − 1 x ( x ) + C ≡ A ( x ) \cdote c o t − 1 x + C = x $
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