Question

log(x+√1+x²)/√1+x²,Integrate it with respect to x.

Answer 1

Answer:

please refer to the attachment

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Answer 2

$\bold{\int\left(\frac{\log (x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}}\right) d x = \frac{[\log (x+\sqrt{x^{2}+1})]^{2}}{2}+c}$

Given:

$\int\left(\frac{\log (x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}}\right) d x$

Step-by-step explanation:

$\text{Let} \ I = \int\left(\frac{\log (x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}}\right) d x$

In this integration, we need to use substitution method.

$\text{Let} \ t = \log (x+\sqrt{1+x^{2}})}$

$d t=\frac{1\left(1+\frac{2 x}{2 \sqrt{1+x^{2}}}\right)}{x+\sqrt{1+x^{2}}} d x$

$\Rightarrow d t=\frac{\left(1+\frac{x}{\sqrt{1+x^{2}}}\right)}{x+\sqrt{1+x^{2}}} d x$

$\Rightarrow d t=\frac{\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}}}{x+\sqrt{1+x^{2}}} d x$

$\Rightarrow d t=\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}} \times \frac{1}{x+\sqrt{1+x^{2}}} d x$

$\therefore d t=\frac{d x}{\sqrt{1+x^{2}}}$

Now, on substituting ‘t’ and ‘dt’ in given equation, we get,

$\Rightarrow I=\int t d t$

$\Rightarrow I=\frac{t^{2}}{2}+C$

Now, on substituting the ‘t’ in above equation, we get,

$\Rightarrow I=\frac{[\log (x+\sqrt{x^{2}+1})]^{2}}{2}+C$

$\therefore \int\left(\frac{\log (x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}}\right) d x = \frac{[\log (x+\sqrt{x^{2}+1})]^{2}}{2}+c$