Question:
If $ \int\frac{dx}{\left(x+2\right)\left(x^{2} +1\right)} = a log\left|1+x^{2}\right| +b tan^{-1} x +\frac{1}{5}log\left|x+2\right|+C$, then
If $ \int\frac{dx}{\left(x+2\right)\left(x^{2} +1\right)} = a log\left|1+x^{2}\right| +b tan^{-1} x +\frac{1}{5}log\left|x+2\right|+C$, then
Updated On: Jul 6, 2022
- $a=\frac{-1}{10} , b=\frac{-2}{5}$
- $a=\frac{1}{10} , b=\frac{-2}{5}$
- $a=\frac{-1}{10} , b=\frac{2}{5}$
- $a=\frac{1}{10} , b=\frac{2}{5}$
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The Correct Option is C
Solution and Explanation
We have, $I = \int\frac{dx}{\left(x+2\right)\left(x^{2}+1\right)}$
Let $ \frac{1}{\left(x+2\right)\left(x^{2}+1\right)} = \frac{A}{x+2} + \frac{Bx+C}{x^{2}+1}$
$ \Rightarrow 1= A\left(x^{2} +1\right) + Bx\left(x+2\right) + C\left(x+2\right) \quad...\left(i\right)$
Put $x = 0$ in $\left(i\right)$, we get $A + 2C = 1$
Put $x= -2$ in $\left(i\right)$, we get $A= \frac{1}{5}$
$\Rightarrow C =\frac{2}{5}$
Put $x= 1$ in $\left(i\right)$, we get
$ 1 = 2A + 3B + 3 C$, we get $B = \frac{-1}{5}$
$ \Rightarrow \int \frac{dx}{\left(x+2\right)\left(x^{2}+1\right)} = \frac{1}{5} \int \frac{dx}{x+2} -\frac{1}{5} \int \frac{xdx}{x^{2}+1} + \frac{2}{5} \int \frac{dx}{x^{2}+1}$
$ = \frac{1}{5 } log\left|x+2\right| -\frac{1}{10 }log\left|x^{2}+1\right| +\frac{2}{5 }tan^{-1} x + C $
Hence, $a= \frac{-1}{10}$ and $b = \frac{2}{5}$
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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.