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Question

(C) Let $I = \int \frac{\cos 2x + x + 1}{x^2 + \sin 2x + 2x} \, dx$. Consider the denominator $t = x^2 + \sin 2x + 2x$. Differentiating with respect t

Vedclass · Accepted Answer

$\frac{1}{2}\log (x^2 + \sin 2x + 2x) + c$

Vedclass · Answer

(C) Let $I = \int \frac{\cos 2x + x + 1}{x^2 + \sin 2x + 2x} \, dx$. Consider the denominator $t = x^2 + \sin 2x + 2x$. Differentiating with respect to $x$,we get $\frac{dt}{dx} = 2x + 2\cos 2x + 2 = 2(x + \cos 2x + 1)$. Thus,$dt = 2(x + \cos 2x + 1) \, dx$,which implies $(x + \cos 2x + 1) \, dx = \frac{1}{2} dt$. Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{1}{2} dt = \frac{1}{2} \int \frac{1}{t} \, dt$. Integrating,we get $I = \frac{1}{2} \log |t| + c$. Substituting back $t$,we get $I = \frac{1}{2} \log |x^2 + \sin 2x + 2x| + c$.

$\int \frac{\cos 2x + x + 1}{x^2 + \sin 2x + 2x} \, dx = $

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