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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int\limits_1^{\sqrt 2 } {\,\,\frac{{{x^2}\,\, + \,\,1}}{{{x^4}\,\, + \,\,1}}} \,dx$ is equal to:
  • A
    $\frac{1}{2}\, \tan^{-1}\, \sqrt 2$
  • ✓
    $\sqrt {\frac{1}{2}}\, \cot^{-1} 2$
  • C
    $\frac{1}{2}\, \tan^{-1} \, \frac{1}{2}$
  • D
    $\sqrt {\frac{1}{2}} \, \tan^{-1} 2$
✓

Answer

Correct option: B.
$\sqrt {\frac{1}{2}}\, \cot^{-1} 2$
b
$\int \frac{x^{2}+1}{x^{4}+1} d x$

divide by $x^{2}$

$\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} d x$

$=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-2+2} d x$

$=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2} d x$

$x-\frac{1}{x}=t$

$1+\frac{1}{x^{2}} d x=d t$

$=\int \frac{1}{t^{2}+2} d t$

$=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+c$

$=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c$

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