Question:
$\int x^{2}\left(ax + b\right)^{-2} dx$ is equal to
$\int x^{2}\left(ax + b\right)^{-2} dx$ is equal to
Updated On: May 13, 2024
- $\frac{2}{a^{2}}\left(x -\frac{b}{a} log\left(ax+b\right)\right)+ C$
- $\frac{2}{a^{2}}\left(x -\frac{b}{a} log\left(ax+b\right)\right) -\frac{x^{2}}{a\left(ax+b\right)}+ C$
- $\frac{2}{a^{2}}\left(x +\frac{b}{a} log\left(ax+b\right)\right) +\frac{x^{2}}{a\left(ax+b\right)}+ C$
- $\frac{2}{a^{2}}\left(x +\frac{b}{a} log\left(ax+b\right)\right) -\frac{x^{2}}{a\left(ax+b\right)}+ C$
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The Correct Option is B
Solution and Explanation
$I= \int\frac{x^{2}}{\left(ax+b\right)^{2}} dx$
Put $ax +b =t \Rightarrow dx =\frac{1}{a} dt $
$\therefore I =\frac{1}{a^{3}}\int\frac{\left(t-b\right)^{2}}{t^{2}} dt =\frac{1}{a^{3}} \int\left(1+\frac{b^{2}}{t^{2}}-\frac{2b}{t}\right) dt$
$= \frac{1}{a^{3}} \left(t -\frac{b^{2}}{t}-2blogt\right) + C $
$ \frac{1}{a^{3}}\left(ax + b -\frac{b^{2}}{ax+b}-2b\, log\left(ax+b\right)\right) + C $
$= \frac{2}{a^{2}} \left(x-\frac{b}{a} log\left(ax+b\right)\right)-\frac{x^{2}}{a\left(ax+b\right)} + C$
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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.