Question:
The value of $\int \limits \frac {(x^2-1)dx}{x^3 \sqrt {2x^4-2x^2+1}} $ is
The value of $\int \limits \frac {(x^2-1)dx}{x^3 \sqrt {2x^4-2x^2+1}} $ is
Updated On: Aug 15, 2022
- $2 \sqrt {2- \frac {2}{x^2}+ \frac {1}{x^4}}+c$
- $2 \sqrt {2+ \frac {2}{x^2}+ \frac {1}{x^4}}+c$
- $\frac {1}{2} \sqrt {2- \frac {2}{x^2}+ \frac {1}{x^4}}+c$
- None of these
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The Correct Option is C
Solution and Explanation
Let I$= \int \limits \frac {(x^2-1)dx}{x^3 \sqrt {2x^4-2x^2+1}}$
$\hspace10mm [dividing \, numerator \, and \, enominator \, by \, x^5]$
$\hspace15mm = \int \limits \frac { \bigg (\frac {1}{x^3} - \frac {1}{x^5} \bigg )dx }{\sqrt {2- \frac {2}{x^2} + \frac {1}{x^4}}} $
Put$2- \frac {2}{x^2}+ \frac {1}{x^4}=t \Rightarrow \bigg ( \frac {4}{x^3}- \frac {4}{x^5} \bigg ) dx=dt$
$\therefore \hspace10mm I= \frac {1}{4} \int \limits \frac {dt}{ \sqrt t}= \frac {1}{4}. \frac {t^{1/2}}{1/2}+c$
$\hspace15mm = \frac {1}{2} \sqrt {2- \frac {2}{x^2}+ \frac {1}{x^4}}+c$
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.