\(\int \sqrt{1+x^2}dx\) is equal to
\(\int \sqrt{1+x^2}dx\) is equal to
\(\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log\mid x+\sqrt{1+x^2}\mid +C\)
\(\frac{2}{3}(1+x^2)^{\frac{2}{3}}+C\)
\(\frac{2}{3}x(1+x^2)^{\frac{3}{2}}+C\)
\(\frac{x^2}{2}\sqrt{1+x^2}+\frac{1}{2}x^2\log\mid x+\sqrt{1+x^2}\mid+C\)
The Correct Option is A
Solution and Explanation
It is known that,\(\int \sqrt{a^2+x^2}dx = \frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\log\mid x+\sqrt{x^2+a^2}\mid+C\)
∴\(\int \sqrt{1+x^2}dx=\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log\mid x+\sqrt{1+x^2}\mid+C\)
Hence, the correct answer is A.
Top Questions on integral
- The value of the integral \( \int_0^1 x^2 \, dx \) is:
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
- Let $ f(x) $ be a positive function and $I_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) dx$ and $I_2 = \int_{-1}^2 f\left(x(1-x)\right) dx.$ Then the value of $\frac{I_2}{I_1}$ is equal to ____
- Evaluate the integral: \[ \int \frac{x^2 + 2x}{\sqrt{x^2 + 1}} \, dx \]
- Evaluate the integral: \[ \int \sqrt{x^2 + 3x} \, dx \]
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(ii) Integrate the function obtained in (i) with respect to \(x\).
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OR
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
