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  • <img alt="\mathrm{The\;integral,\;\;I=\int \frac{\left(x^{2}-3 x+1\right)}{\sqrt{-x^{2}+1}}dx\;\;is\;equals \;to}" src="https://learn.careers360.com/latex-image/?%5Cmathrm%7BThe%5C%3Bintegral%2C%5C

\mathrm{The\;integral,\;\;I=\int \frac{\left(x^{2}-3 x+1\right)}{\sqrt{-x^{2}+1}}dx\;\;is\;equals \;to}

Option: 1

\frac{(x-6) \sqrt{1-x^{2}}-3 \arcsin (x)}{2}+C


Option: 2

-\frac{(x-6)^2 \sqrt{1-x^{2}}-3 \arcsin (x)}{2}+C


Option: 3

-\frac{(x-6) \sqrt{1-x^{2}}+3 \arcsin (x)}{2}+C


Option: 4

-\frac{(x-6) \sqrt{1-x^{2}}-3 \arcsin (x)}{2}+C


Answers (1)

 

 

Application of Special Integral Formula (Part 2) -

Integration of the type

\\\mathrm{1.\;\;\int \frac{a x^{2}+b x+c}{\left(p x^{2}+q x+r\right)} d x}\\\\\mathrm{2.\;\;\int \frac{\left(a x^{2}+b x+c\right)}{\sqrt{p x^{2}+q x+r}} d x}\\\\\mathrm{3.\;\;\int\left(a x^{2}+b x+c\right) \sqrt{p x^{2}+q x+r} d x}

\\\text{Substitute,}\\\\\mathrm{a x^{2}+b x+c=\lambda\left(p x^{2}+q x+r\right)+\mu\left\{\frac{d}{d x}\left(p x^{2}+q x+r\right)\right\}+\gamma}

Find λ, μ and γ. These integrations reduces to integration of three independent functions.

 

Integration of the form \mathbf{\int \frac{k(x)}{ax^2+bx+c}\;dx}

here, k(x) is a polynomial of degree greater than 2

To solve this type of integration, divide the numerator by the denominator and express the intagral as 

\mathrm{Q(x)+\frac{R(x)}{ax^2+bx+c}} 

Here, R(x) is a linear function of x. 

-

 

 

\\\text { Let } x^{2}-3 x+1=A\left(1-x^{2}\right)+B \frac{d}{d x}\left(1-x^{2}\right)+C\\\text{Comparing the coefficients like powers of x}\\\mathrm{A=-1,\;B=3/2,\;and\;\;C=2}\\\\\int \frac{\mathrm{x}^{2}-3 \mathrm{x}+1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\int \frac{-\left(1-\mathrm{x}^{2}\right)+\frac{3}{2}(-2 \mathrm{x})+2}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{d} \mathrm{x}\\\\=-\int \sqrt{1-\mathrm{x}^{2}} \mathrm{d} \mathrm{x}-\int \frac{3 \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{d} \mathrm{x}+2 \int \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}

\\=-\left[\frac{\mathrm{x} \sqrt{1-\mathrm{x}^{2}}}{2}+\frac{1}{2} \sin ^{-1} \mathrm{x}\right]-3\left(-\frac{1}{2}\right) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x+2 \int \frac{1}{\sqrt{1-x^{2}}} d x\\\\=-\frac{x}{2} \sqrt{1-x^{2}}+\frac{3}{2} \sin ^{-1} x+3 \sqrt{1-x^{2}}+C\\\\=\frac{6-x}{2} \sqrt{1-x^{2}}+\frac{3}{2} \sin ^{-1} x+C\\\\=-\frac{(x-6) \sqrt{1-x^{2}}-3 \arcsin (x)}{2}+C

Posted by

Sumit Saini

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