Question

Prove that (i) \(\int_{1}^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}+\sqrt{\mathrm{x}^{2}+1}\right)^{\mathrm{n}}}=\frac{\mathrm{n}}{\mathrm{n}^{2}-1}, \mathrm{n}>1\) (ii) \(\int_{1}^{\infty} \frac{d x}{\left(1+e^{x}\right)\left(1+e^{-x}\right)}=1\) (iii) \(\int_{0}^{\infty} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} d x=0\) (iv) \(\int_{0}^{\infty} \frac{\sqrt{x}}{(1+x)^{2}} d x=\frac{1}{2}+\frac{1}{4} \pi\).

Short answer

Question: Evaluate the following integral and provide the final solution: $$\int_{1}^{\infty} \frac{dx}{\left(1+e^{x}\right)\left(1+e^{-x}\right)}$$ Answer: To evaluate this integral, first combine the denominator terms as follows: $$\int_{1}^{\infty} \frac{dx}{\left(1+e^{x}\right)\left(1+e^{-x}\right)} = \int_{1}^{\infty} \frac{dx}{2 + e^x + e^{-x}}$$ Calculate the antiderivative: $$\int_{1}^{\infty} \frac{dx}{2 + e^x + e^{-x}} = \left[-\ln|e^x - 1|\right]_1^{\infty}$$ Finally, apply the limits and find the solution: $$-\ln|e^\infty - 1| + \ln|e^1 - 1| = -\infty + \ln2$$ Since the limit doesn't converge, the integral has no finite solution.

Step-by-step solution

(Integral 1: Substitution)

Let \(u = x + \sqrt{x^2 + 1}\), then the substitution can be rewritten as \(x = u - \sqrt{u^2 - 1}\). Differentiating with respect to x, we get \(\frac{dx}{du} = 1 + \frac{u}{\sqrt{u^2 - 1}}\). The integral now becomes, $$\int_{1}^{\infty} \frac{1 + \frac{u}{\sqrt{u^2 - 1}}}{u^n}du$$ Now we can split the integral into two parts, which are much easier to evaluate. Apply the substitution bounds, and consider that \(n > 1\), we have, $$\frac{\mathrm{n}}{\mathrm{n}^{2}-1}=\int_{2}^{\infty} \frac{1}{u^n}du + \int_{2}^{\infty} \frac{1}{u^{n-1}\sqrt{u^2 - 1}}du$$ Evaluate each integral and combine the results to find the solution.

(Integral 2: Combine denominator terms)

Observe that the denominator can be combined to form a more simplified expression, $$\int_{1}^{\infty} \frac{dx}{\left(1+e^{x}\right)\left(1+e^{-x}\right)} = \int_{1}^{\infty} \frac{dx}{2 + e^x + e^{-x}}$$ Proceed to evaluate the integral with the simplified expression.

(Integral 3: Integration by parts)

To evaluate the integral, use integration by parts with \(u = \ln x\) and \(dv = \frac{x}{(1+x^2)^2} dx\). On differentiating and integrating, we get, $$ du = \frac{1}{x} dx \\ v = \int \frac{x}{(1+x^2)^2} dx $$ Apply integration by parts, $$\int_{0}^{\infty} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} d x = [uv]_{0}^{\infty} - \int_{0}^{\infty} v du$$ Evaluate the RHS of the equation to find the final solution for the integral.

(Integral 4: Trigonometric substitution)

To evaluate the integral, we can use the trigonometric substitution \(x = \tan \theta\), such that \(dx = \sec^2\theta d\theta\) and \(\sqrt{x} = \sqrt{\tan \theta}\). The integral now becomes, $$\int_{0}^{\infty} \frac{\sqrt{x}}{(1+x)^{2}} d x = \int_{0}^{\pi/2} \frac{\sqrt{\tan \theta}\sec^2\theta}{(1 + \tan \theta)^2}d \theta$$ Proceed to evaluate the integral and convert the bounds back to the original variable x. The final solution for the integral is \(\frac{1}{2}+\frac{1}{4}\pi\).