Question:
$ \int tan^{-1} \sqrt{x}\,dx $ is equal to
$ \int tan^{-1} \sqrt{x}\,dx $ is equal to
Updated On: Jul 6, 2022
- $ \left(x + 1\right)tan^{-1}\sqrt{x} -\sqrt{x} + C $
- $ x\,tan^{-1}\sqrt{x} -\sqrt{x} + C $
- $ \sqrt{x} - x\, tan^{-1}\sqrt{x} + C $
- $ \sqrt{x} - (x+1) tan^{-1}\sqrt{x} + C $
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The Correct Option is A
Solution and Explanation
We have, $I = \int 1\cdot tan^{-1}\sqrt{x}\,dx$
$ \Rightarrow I= tan^{-1}\sqrt{x} \cdot\left(x\right) -\int\frac{1}{ 1+x} \times \frac{1}{2\sqrt{x}} \times x \, dx $
$ = x\,tan^{-1}\sqrt{x}-\int\frac{x}{ \left(1+x\right) \,2\,\sqrt{x}} dx$
$ = x\, tan^{-1} \sqrt{x} -\int\left(\frac{1+x}{\left(1+x\right)\, 2\,\sqrt{x}} - \frac{1}{\left(1+x\right)\,2\,\sqrt{x}}\right) dx$
$= x\, tan^{-1} \sqrt{x}- \int\frac{dx}{2\,\sqrt{x}} + \int \frac{dx}{2\,\sqrt{x} \left(1+x\right)} $
$= x\,tan^{-1}\sqrt{x} - \sqrt{x} + tan^{-1}\sqrt{x} +C $
$ =\left(x+1\right)tan^{-1} \sqrt{x} - \sqrt{x} + 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.