JEE Main 2019 — Mathematics Calculus

We have $\displaystyle I=\int \frac{dx}{{({x}^{2}-2x+10)}^{2}}=\int \frac{dx}{{[{(x-1)}^{2}+{3}^{2}]}^{2}}$I=∫(x2−2x+10)2dx​=∫[(x−1)2+32]2dx​

Let $x –1=3\mathrm{tan}\theta \Rightarrow dx=3{\mathrm{sec}}^{2}\theta d\theta$x–1=3tanθ⇒dx=3sec2θdθ

$$\Rightarrow I=\int \frac{3se{c}^{2}\theta d\theta }{81se{c}^{4}\theta }=\frac{1}{27}\int co{s}^{2}\theta d\theta$$⇒I=∫81sec4θ3sec2θdθ​=271​∫cos2θdθ

$$=\frac{1}{54}\int (1+\mathrm{cos}2\theta )d\theta$$=541​∫(1+cos2θ)dθ

$$=\frac{1}{54}(\theta +\frac{\mathrm{sin}2\theta }{2})+C$$=541​(θ+2sin2θ​)+C

$=\frac{1}{54}(\theta +\mathrm{sin}\theta \mathrm{cos}\theta )+C$=541​(θ+sinθcosθ)+C, where $C$C is the constant of integration.

$$=\frac{1}{54}[{\mathrm{tan}}^{-1}(\frac{x-1}{3})+\frac{(\frac{x-1}{3})}{1+{(\frac{x-1}{3})}^{2}}]+C$$=541​[tan−1(3x−1​)+1+(3x−1​)2(3x−1​)​]+C

$$=\frac{1}{54}[ta{n}^{-1}(\frac{x-1}{3})+\frac{3(x-1)}{{x}^{2}-2x+10}]+C$$=541​[tan−1(3x−1​)+x2−2x+103(x−1)​]+C

Hence, A=\frac{1}{54} & f(x)=3(x-1)