Question

Evaluate the following integrals: (i) \(\int \frac{\sqrt{x^{4}+x^{-4}+2}}{x^{3}} d x\) (ii) \(\int \frac{d x}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x\) (iii) \(\int \frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}} d x\) (iv) \(\int\left(\frac{1-x^{-2}}{x^{1 / 2}-x^{-1 / 2}}-\frac{2}{x^{3 / 2}}+\frac{x^{-2}-x}{x^{1 / 2}-x^{-1 / 2}}\right) d x\)

Short answer

Question: Evaluate the following integrals: (i) \(\int \frac{\sqrt{x^{4}+x^{-4}+2}}{x^{3}} d x\) (ii) \(\int \frac{dx}{(\sqrt{2x-3}+\sqrt{2x+3})}\) (iii) \(\int \frac{(x^{2}-\sqrt{x}) (\sqrt{x}+1)}{x \sqrt{x}+x+\sqrt{x}} dx\) (iv) \(\int\left[\frac{1 - x^{-2}}{x^{1/2}-x^{-1/2}}-\frac{2}{x^{3/2}}\right] d x\) Solution: (i) The integral does not have a general solution, so the result stands as \(\int \sqrt{1+\sec^4t} dt\). (ii) The result of the integral is \(\ln(|\sqrt{2x-3} - \sqrt{2x+3}|^2+2) + C\). (iii) The result of the integral is \(\frac{x^3}{3} - 2x^{3/2} + 2x^{1/2} - ln|x| + C\). (iv) The result of the integral is \(\frac{x^2}{2} - x - 4x^{1 / 2} + C\).

Step-by-step solution

(i) Rewrite the integrand

For the first integral, rewrite the expression in the square root by factoring out a common factor: \(\int \frac{\sqrt{x^{4}+x^{-4}+2}}{x^{3}} d x = \int \frac{\sqrt{(x^{4}+1)(x^{-4}+1)}}{x^{3}} d x\)

(i) Recognize a trigonometric substitution

Recognize that the expression becomes simpler if we use the trigonometric substitution \(x = \tan{t}\). Taking the derivative with respect to t gives us: \(dx = \sec^{2}{t} dt\). Applying the substitution, the integral becomes: \(\int \frac{\sqrt{\tan^{4}{t}+\sec^{4}{t}}}{\tan^{3}{t}}\sec^{2}{t} dt\)

(i) Simplify the integral

Simplify the integral by canceling out common terms and by applying the trigonometric identity \(\tan^2{t}+\sec^2{t}=1+\sec^2{t}\): \(\int \sqrt{1+\sec^4t}dt\) We don't have a general solution for this integral, so the result stands as \(\int \sqrt{1+\sec^4t} dt\)

(ii) Make a substitution

For the second integral, we can simplify by making a substitution \(u = \sqrt{2x-3} - \sqrt{2x+3}\), which tells us that \( du = \frac{2 dx}{\sqrt{2x-3}+\sqrt{2x+3}}\), so \(dx = \frac{1}{2}du(\sqrt{2x-3}+\sqrt{2x+3})\). Substituting the integral becomes: \(\int 2\frac{u}{u^2+2}du\)

(ii) Integrate the expression

Integrate the above expression to find the result: \(\int 2\frac{u}{u^2+2}du = \ln(|u^2+2|) + C\) The integrating factor is \(u\), so we have: \(\ln(|\sqrt{2x-3} - \sqrt{2x+3}|^2+2) + C\)

(iii) Simplify the expression

For the third integral, we should simplify the expression by expanding the numerator and looking for common factors: \(\int \frac{(x^{2}-\sqrt{x}) (\sqrt{x}+1)}{x \sqrt{x}+x+\sqrt{x}} dx = \int \frac{x^3 + x^2 - x\sqrt{x} - \sqrt{x}}{x^{3/2}+x\sqrt{x} +x} dx\)

(iii) Integrate term by term

Integrate the simplified expression term by term: \(\int (x^2 - x^{1/2} + x^{-1/2} - x^{-1}) dx = \frac{x^3}{3} - 2x^{3/2} + 2x^{1/2} - ln|x| + C\)

(iv) Combine the fractions

For the fourth integral, combine the fractions: $\int\left(\frac{(1-x^{-2})-(x^{-2}-x)}{x^{1 / 2}-x^{-1 / 2}}-\frac{2}{x^{3 / 2}}\right) d x$ $= \int\left(\frac{(1-x)(x^2-1)}{(x^{1 / 2}-x^{-1 / 2})(x^2-1)}-\frac{2}{x^{3 / 2}}\right) dx$ $= \int\left(\frac{x^2-x}{x^{3 / 2}-x^{1 / 2}}-\frac{2}{x^{3 / 2}}\right) dx$

(iv) Simplify and integrate

Simplify the expression by taking out the common factor and integrate term by term: \(\int\left(x - 1 - \frac{2}{x^{1 / 2}}\right) dx = \frac{x^2}{2} - x - 4x^{1 / 2} + C\)