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Chapter 1: Problem 6

Evaluate the following integrals : $$\int \frac{x d x}{x-\sqrt{x^{2}-1}}$$

Short Answer

Expert verified
Question: Evaluate the integral of the function $\frac{x}{x-\sqrt{x^{2}-1}}$ with respect to $x$. Answer: $\int \frac{x dx}{x-\sqrt{x^{2}-1}} = \text{arccosh}^2(x) - x\sqrt{x^2 - 1} + C$

Step by step solution

01

Perform Substitution

Let us perform the substitution: $$x = \cosh(u)$$ In this case, we have \(\text{d}x = \sinh(u) \text{d}u\), and the integral becomes: $$\int \frac{\cosh(u)\sinh(u) du}{\cosh(u) - \sqrt{\cosh^2(u)-1}}$$ Now, we know that $$\sinh^2(u) + 1 = \cosh^2(u)$$, so: $$\sqrt{\cosh^2(u) - 1} = \sinh(u)$$ Now, our integral becomes: $$\int \frac{\cosh(u)\sinh(u) du}{\cosh(u) - \sinh(u)}$$
02

Simplify the Integral

Next, factor out a \(\cosh(u)\) from the denominator: $$\int \frac{\cosh(u)\sinh(u) du}{\cosh(u)(1 - \tanh(u))}$$ Since \(\tanh(u) = \frac{\sinh(u)}{\cosh(u)}\), our integral becomes: $$\int \frac{\cosh(u)\sinh(u) du}{\cosh(u)(1 - \frac{\sinh(u)}{\cosh(u)})}$$ Now, cancel out a \(\cosh(u)\) from the numerator and denominator: $$\int \frac{\sinh(u) du}{1 - \frac{\sinh(u)}{\cosh(u)}}$$
03

Use Integration by Parts

Use integration by parts with: $$u = \sinh(u) \Rightarrow \text{d}u = \cosh(u) du$$ $$\text{d}v = du \Rightarrow v = u$$ Now, our integral becomes: $$uv - \int v \text{d}u = u^2 - \int u \cosh(u) du$$
04

Perform Second Integration by Parts

Again, we apply integration by parts to the remaining integral: $$u_1 = u \Rightarrow du_1 = du$$ $$dv_1 = \cosh(u) du \Rightarrow v_1 = \sinh(u)$$ Now, the integral becomes: $$u^2 - (u_1 v_1 - \int v_1 du_1) = u^2 - (u\sinh(u) - \int \sinh(u) du)$$ We know that $$\int \sinh(u) du = \cosh(u) + C$$, so our integral becomes: $$u^2 - u\sinh(u) + \cosh(u) + C$$
05

Substitute Back

Finally, substitute back with \(x = \cosh(u)\), we get: $$u = \text{arccosh}(x)$$ So, the final result is: $$\text{arccosh}^2(x) - x\sqrt{x^2 - 1} + C$$ Therefore, $$\int \frac{x dx}{x-\sqrt{x^{2}-1}} = \text{arccosh}^2(x) - x\sqrt{x^2 - 1} + C$$

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