\(∫^1_0\dfrac{x}{x^2-4}dx=?\)
\(\dfrac{-2\pi}{6}\)
\(\ln \frac{\sqrt{3}}{2}\)
\(ln(\dfrac{3}{2})\)
\(ln(\dfrac{√3}{2})\)
\(ln(\dfrac{3}{√2})\)
The Correct Option is B
Solution and Explanation
The integral we need to evaluate is:
\[ \int_0^1 \frac{x}{x^2 - 4} dx \]
Let \( u = x^2 - 4 \). Then \( du = 2x dx \), and \( x dx = \frac{1}{2} du \).
When \( x = 0 \), \( u = 0^2 - 4 = -4 \).
When \( x = 1 \), \( u = 1^2 - 4 = -3 \).
The integral becomes:
\[ \int_{-4}^{-3} \frac{1}{2u} du = \frac{1}{2} \int_{-4}^{-3} \frac{1}{u} du = \frac{1}{2} [\ln |u|]_{-4}^{-3} \]
\[ = \frac{1}{2} (\ln|-3| - \ln|-4|) = \frac{1}{2} (\ln 3 - \ln 4) = \frac{1}{2} \ln \frac{3}{4} = \ln \sqrt{\frac{3}{4}} = \ln \frac{\sqrt{3}}{2} \]
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C