Question:
$\int\left\{\frac{\left(log \,x -1\right)}{1+\left(log \,x\right)^{2}}\right\}^{2}dx$ is equal to :
$\int\left\{\frac{\left(log \,x -1\right)}{1+\left(log \,x\right)^{2}}\right\}^{2}dx$ is equal to :
Updated On: Jul 28, 2022
- $\frac{x}{\left(log\,x\right)^{2}+1}+c$
- $\frac{xe^{x}}{1+x^{2}}+c$
- $\frac{x}{x^{2} +1}+c$
- $\frac{log\,x}{\left(log\,x\right)^{2} +1}+c$
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The Correct Option is A
Solution and Explanation
Let $I = \int \left(\frac{log\,x-1}{ 1+\left(log\,x\right)^{2}}\right)^{2} dx$
$= \int \frac{\left(log\,x\right)^{2}+1-2log\,x}{\left(1+\left(log\,x\right)^{2}\right)^{2}}dx$
$= \int \frac{dx}{1+\left(log\,x\right)^{2}}- \int \frac{2log\,x}{\left(1+\left(log\,x\right)^{2}\right)^{2}}dx$
$= \frac{x}{1+\left(log\,x\right)^{2}}+\int \frac{2log\,x}{\left(1+\left(log\,x\right)^{2}\right)^{2}}dx$
$-\int \frac{2log\,x}{\left(1+\left(log\,x\right)^{2}\right)^{2}}$
$= \frac{x}{1+\left(log\,x\right)^{2}}+c$
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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.