int frac{x + 1}{sqrt{1 + x^2}} dx = | vedclass

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Question

(C) We need to evaluate the integral $I = \int \frac{x + 1}{\sqrt{x^2 + 1}} dx$. Split the integral into two parts: $I = \int \frac{x}{\sqrt{x^2 + 1}}

Vedclass · Accepted Answer

$\sqrt{1 + x^2} + \log \{ x + \sqrt{1 + x^2} \} + c$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int \frac{x + 1}{\sqrt{x^2 + 1}} dx$. Split the integral into two parts: $I = \int \frac{x}{\sqrt{x^2 + 1}} dx + \int \frac{1}{\sqrt{x^2 + 1}} dx$. For the first part,let $t = x^2 + 1$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$. So,$\int \frac{x}{\sqrt{x^2 + 1}} dx = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} = \sqrt{t} = \sqrt{x^2 + 1}$. For the second part,we use the standard integral formula $\int \frac{1}{\sqrt{x^2 + a^2}} dx = \log |x + \sqrt{x^2 + a^2}| + c$. Thus,$\int \frac{1}{\sqrt{x^2 + 1}} dx = \log |x + \sqrt{x^2 + 1}| + c$. Combining both parts,we get $I = \sqrt{x^2 + 1} + \log |x + \sqrt{x^2 + 1}| + c$.

$\int \frac{x + 1}{\sqrt{1 + x^2}} dx = $

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