The value of ∫loge⁡x+x2+1x2+1dx, is - Infinity Learn by Sri Chaitanya

A
$2 \log_{e} (x + \sqrt{x^{2} + 1}) + C$
B
${\{\log_{e} (x + \sqrt{x^{2} + 1})\}}^{2} + C$
C
$\log (x + \sqrt{x^{2} + 1}) + C$
D
none of these

Solution:

We have.

$\begin{array}{l} I = \int \frac{\log_{c} (x + \sqrt{x^{2} + 1})}{\sqrt{x^{2} + 1}} d x ∣ \\ \Rightarrow l = \int \log_{c} (x + \sqrt{x^{2} + 1}) d \{\log_{e} (x + \sqrt{x^{2} + 1})\} \\ \Rightarrow I = \frac{1}{2} {\{\log_{p} (x + \sqrt{x^{2} + 1})\}}^{2} + C \end{array}$

The value of $\int \frac{\log_{e} (x + \sqrt{x^{2} + 1})}{\sqrt{x^{2} + 1}} d x$ , is

Solution:

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