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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
The primitive of the function $\text{f(x)}=\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}},\text{ a} > 0$ is :
  • ✓
    $\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$
  • B
    $\log_\text{e}\text{a}\cdot\text{a}^{\text{x}+\frac{1}{\text{x}}}$
  • C
    $\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\text{x}}{\log_\text{e}\text{a}}$
  • D
    $\text{x}\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$
✓

Answer

Correct option: A.
$\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$
$\text{f(x)}=\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}}$
$\therefore\ \int\text{f(x)}\text{dx}=\int\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}}$
Let $\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{t}$
$\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\ \int\text{f(x)}\text{dx}=\int\text{a}^{\text{t}}\text{ dt}$
$=\frac{\text{a}^{\text{t}}}{\log_\text{e}\text{a}}+\text{C}$
$=\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}+\text{C}$
$\Big(\because\text{t}=\text{x}+\frac{1}{\text{x}}\Big)$

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