\(\int \sqrt{1+x^2}dx\) is equal to
\(\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log\mid x+\sqrt{1+x^2}\mid +C\)
\(\frac{2}{3}(1+x^2)^{\frac{2}{3}}+C\)
\(\frac{2}{3}x(1+x^2)^{\frac{3}{2}}+C\)
\(\frac{x^2}{2}\sqrt{1+x^2}+\frac{1}{2}x^2\log\mid x+\sqrt{1+x^2}\mid+C\)
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.
If \[ \int e^x (x^3 + x^2 - x + 4) \, dx = e^x f(x) + C, \] then \( f(1) \) is:
The value of : \( \int \frac{x + 1}{x(1 + xe^x)} dx \).
A certain reaction is 50 complete in 20 minutes at 300 K and the same reaction is 50 complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction. Given: \[ R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}, \quad \log 4 = 0.602 \]
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.
These are tabulated below along with the meaning of each part.