Question

Evaluate the following integrals: (i) \(\int \frac{d x}{x \sqrt{25 x^{2}-2}}\) (ii) \(\int \frac{\mathrm{dx}}{(\mathrm{x}+1) \sqrt{\mathrm{x}^{2}+2 \mathrm{x}}}\) (iii) \(\int \frac{d x}{(2 x-1) \sqrt{(2 x-1)^{2}-4}}\)

Short answer

Question: Find the antiderivatives of the following functions: (i) \(\int \frac{d x}{x \sqrt{25 x^{2}-2}}\) (ii) \(\int \frac{dx}{(x+1) \sqrt{x^2+2x}}\) (iii) \(\int \frac{dx}{(2x-1) \sqrt{(2x-1)^2-4}}\) Answer: (i) \(\frac{1}{25} \ln(\sqrt{25x^2 - 2}) + C\) (ii) \(\ln(\sqrt{(x+1)^2 - 1}) + C\) (iii) \(-\frac{1}{2}\ln(\sqrt{(2x-1)^2 - 4}) + C\)

Step-by-step solution

Substitution

Let \(u^2 = 25x^2 - 2\). Then \(2u \mathrm{d}u = 50x \mathrm{d}x\). Now, substitute \(u\) and \(\mathrm{d}u\) into the integral: \(\int \frac{1}{x \sqrt{25x^{2}-2}} \mathrm{d}x = \int \frac{2u\mathrm{d}u}{50x^2u}\)

Simplify and Integrate

Simplify the integral: \(\int \frac{1}{25x} \mathrm{d}u\) Now, integrate with respect to u: \(\frac{1}{25} \int \frac{1}{x} \mathrm{d}u = \frac{1}{25} \ln(u) + C\)

Substitute back for x

Finally, substitute x back into the expression: \(\frac{1}{25} \ln(\sqrt{25x^2 - 2}) + C\) (ii) \(\int \frac{dx}{(x+1) \sqrt{x^2+2x}}\)

Substitution

Complete the square in the square root: \(x^2 + 2x = (x+1)^2 - 1\). Let \(u^2 = (x+1)^2 - 1\). Then \(2u \mathrm{d}u = 2(x+1) \mathrm{d}x\). Now, substitute \(u\) and \(\mathrm{d}u\) into the integral: \(\int \frac{1}{(x+1) \sqrt{x^2+2x}} \mathrm{d}x = \int \frac{2u\mathrm{d}u}{2(x+1) u}\)

Simplify and Integrate

Simplify the integral: \(\int \frac{1}{x+1} \mathrm{d}u\) Now, integrate with respect to u: \(\ln(u) + C\)

Substitute back for x

Finally, substitute x back into the expression: \(\ln(\sqrt{(x+1)^2 - 1}) + C\) (iii) \(\int \frac{dx}{(2x-1) \sqrt{(2x-1)^2-4}}\)

Substitution

Let \(u^2 = (2x-1)^2 - 4\). Then \(2u \mathrm{d}u = 4(2x-1) \mathrm{d}x\). Now, substitute \(u\) and \(\mathrm{d}u\) into the integral: \(\int \frac{1}{(2x-1) \sqrt{(2x-1)^2-4}} \mathrm{d}x = \int \frac{4u\mathrm{d}u}{4(1-2x)u}\)

Simplify and Integrate

Simplify the integral: \(\int \frac{1}{1-2x} \mathrm{d}u\) Now, integrate with respect to u: \(-\frac{1}{2}\ln(u) + C\)

Substitute back for x

Finally, substitute x back into the expression: \(-\frac{1}{2}\ln(\sqrt{(2x-1)^2 - 4}) + C\)