NDA 2017
Exam NDA
Year 2017
Subject Mathematics
Chapter Indefinite Integration NDA 2017 · Mathematics · Indefinite Integration
∫(In x)^ -1 d x-∫(In x)^ -2 d x is equal to
∫ ( In x ) − 1 d x − ∫ ( In x ) − 2 d x \int(\operatorname{In} x)^{-1} d x-\int(\operatorname{In} x)^{-2} d x ∫ ( In x ) − 1 d x − ∫ ( In x ) − 2 d x is equal to
Options A x ( In x ) − 1 + c x(\operatorname{In} x)^{-1}+c x ( In x ) − 1 + c B x ( In x ) − 2 + c x(\operatorname{In} x)^{-2}+c x ( In x ) − 2 + c C x ( In x ) + c x(\operatorname{In} x)+c x ( In x ) + c D x ( In x ) 2 + c x(\operatorname{In} x)^2+c x ( In x ) 2 + c Correct answer A. x ( In x ) − 1 + c x(\operatorname{In} x)^{-1}+c x ( In x ) − 1 + c
Step-by-step solution
∫ ( ln x ) − 1 d x − ∫ ( ln x ) − 2 d x = ∫ [ 1 ln x − 1 ( ln x ) 2 ] d x \begin{aligned} & \int(\ln x)^{-1} d x-\int(\ln x)^{-2} d x \\ & =\int\left[\frac{1}{\ln x}-\frac{1}{(\ln x)^2}\right] d x\end{aligned} ∫ ( ln x ) − 1 d x − ∫ ( ln x ) − 2 d x = ∫ [ ln x 1 − ( ln x ) 2 1 ] d x Put ℓ n x = t ⇒ x = e t d x = e t . d t . \begin{aligned} & \text{ Put } \ell \mathrm{n} \mathrm{x}=\mathrm{t} \Rightarrow \mathrm{x}=\mathrm{e}^{\mathrm{t}} \\ & \mathrm{dx}=\mathrm{e}^{\mathrm{t}} . \mathrm{dt} .\end{aligned} Put ℓ nx = t ⇒ x = e t dx = e t . dt . ∴ ∫ [ 1 ln x − 1 ( ln x ) 2 ] d x = ∫ ( 1 t − 1 t 2 ) ⋅ e t ⋅ d t = ∫ e t ⋅ ( 1 t − 1 t 2 ) ⋅ d t \begin{aligned} & \therefore \int\left[\frac{1}{\ln \mathrm{x}}-\frac{1}{(\ln \mathrm{x})^2}\right] \mathrm{dx}=\int\left(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}^2}\right) \cdot \mathrm{e}^{\mathrm{t}} \cdot \mathrm{dt} \\ & =\int \mathrm{e}^{\mathrm{t}} \cdot\left(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}^2}\right) \cdot \mathrm{dt}\end{aligned} ∴ ∫ [ ln x 1 − ( ln x ) 2 1 ] dx = ∫ ( t 1 − t 2 1 ) ⋅ e t ⋅ dt = ∫ e t ⋅ ( t 1 − t 2 1 ) ⋅ dt = ∫ e t ⋅ ( 1 t − 1 t 2 ) ⋅ d t = e t t + c = x ln x + c \begin{aligned} & =\int \mathrm{e}^{\mathrm{t}} \cdot\left(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}^2}\right) \cdot \mathrm{dt} \\ & =\frac{\mathrm{e}^{\mathrm{t}}}{\mathrm{t}}+\mathrm{c}=\frac{\mathrm{x}}{\ln \mathrm{x}}+\mathrm{c}\end{aligned} = ∫ e t ⋅ ( t 1 − t 2 1 ) ⋅ dt = t e t + c = ln x x + c =x(\ln x)^{-1}+c$
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