Question:
Evaluate: $\int\left\{\left(2x-3\right)^{5}+\frac{1}{\left(7x-5\right)^{3}}+\frac{1}{\sqrt{5x-4}}+\frac{1}{2-3x}+\sqrt{3x+2}\right\}dx$
Evaluate: $\int\left\{\left(2x-3\right)^{5}+\frac{1}{\left(7x-5\right)^{3}}+\frac{1}{\sqrt{5x-4}}+\frac{1}{2-3x}+\sqrt{3x+2}\right\}dx$
Updated On: Jul 6, 2022
- $I= \frac{1}{12}\left(2x-3\right)^{6}-\frac{1}{14}\left(7x-5\right)^{-2}+\frac{2}{5}\sqrt{5x-4}-\frac{1}{3}log\left|2-3x\right|+\frac{2}{9}\left(3x+2\right)^{\frac{3}{2}}+C$
- $I= \frac{1}{12}\left(2x-3\right)^{6}-\frac{1}{14}\left(7x-5\right)^{-2}+\frac{2}{5}\sqrt{5x-4}$
- $I= \frac{1}{12}\left(2x-3\right)^{4}+\frac{1}{14}\left(7x-5\right)^{-2}$
- None of these
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The Correct Option is A
Solution and Explanation
$\int\left\{\left(2x-3\right)^{5}+\frac{1}{\left(7x-5\right)^{3}}+\frac{1}{\sqrt{5x-4}}+\frac{1}{2-3x}+\sqrt{3x+2}\right\}dx$
$\Rightarrow I= \int\left(2x-3\right)^{5}dx +\int\left(7x-5\right)^{-3}dx +\int \left(5x-4\right)^{-\frac{1}{2}}dx +\int\frac{1}{2-3x}dx +\int \sqrt{3x+2}dx$
$\Rightarrow I = \frac{\left(2x-3\right)^{6}}{2\times6} + \frac{\left(7x - 5\right)^{-2}}{7\times\left(-2\right)} + \frac{\left(5x-4\right)^{\frac{1}{2}}}{5\times\frac{1}{2}} + \left(\frac{1}{-3}\right)log \left|2-3x\right| + \frac{\left(3x+2\right)^{\frac{3}{2}}}{3\times\frac{3}{2}}+C$
$\Rightarrow I = \frac{1}{12} \left(2x-3\right)^{6} - \frac{1}{14} \left(7x-5\right)^{-2}+\frac{2}{5}\sqrt{5x - 4}-\frac{1}{3}log \left|2-3x\right| + \frac{2}{9}\left(3x+2\right)^{\frac{3}{2}} + C$
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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.