Question

Evaluate the following integrals : (i) \(\int_{1}^{\infty} \frac{d x}{x^{2}(x+1)}\) (ii) \(\int_{0}^{\infty} x^{3} e^{-x^{2}} d x\) (iii) \(\int_{0}^{\frac{1}{6}} \frac{\mathrm{dx}}{\mathrm{x} \ln ^{2} \mathrm{x}}\) (iv) \(\int_{-\infty}^{\infty} \frac{d x}{x^{2}+2 x+2}\)

Short answer

Answer: We can use the trigonometric substitution \(x+1 = \tan{u}\) to evaluate this integral.

Step-by-step solution

Write the integrand as partial fractions

We want to write the integrand as partial fractions: \(\frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}\)

Eliminate the denominators

By eliminating the denominators, we get \(1 = A(x^2)(x+1) + B(x)(x+1) + C(x^2)(x)\). Now we can find A, B, and C.

Find A, B, and C

By setting x = 0, we get \(A = 1\), and by setting x = -1, we get \(C = -1\). Substituting these values back into the equation above, we get \(1 = x^3 - x^2 + Bx(x+1)\). Comparing the coefficients, we get \(B = -1\). The partial fractions are \(\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x+1}\).

Evaluate the integral

Now we can rewrite our integral as \(\int_{1}^{\infty} \left( \frac{1}{x} - \frac{1}{x^2} - \frac{1}{x+1} \right) dx\). This can be calculated as a limit: \(\lim_{b \to \infty} \int_{1}^{b} \left( \frac{1}{x} - \frac{1}{x^2} - \frac{1}{x+1} \right) dx = \lim_{b \to \infty} [\ln{x} + \frac{1}{x} - \ln{(x+1)}]_{1}^{b}\). As \(b \to \infty\), the limit becomes \(- \ln{2} + 1\). So, the integral is equal to \(- \ln{2} + 1\). (ii) \(\int_{0}^{\infty} x^{3} e^{-x^{2}} d x\)

Use substitution method

Let \(u = x^2\), so \(\frac{1}{2} du = x dx\). Our integral becomes \(\frac{1}{2} \int_{0}^{\infty} u^{\frac{3}{2}-1} e^{-u} du\).

Evaluate the integral

We use the formula for the gamma function, which is \(\Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} dx\). We have \(n = \frac{5}{2}\). The gamma function for positive half-integers can be expressed as \(\Gamma\left(\frac{2k+1}{2}\right) = \frac{\sqrt{\pi}(2k-1)!!}{2^k}\), where \(!!\) represents the double factorial. Therefore, substituting the value of n and simplifying, we get \(\frac{3}{4}\sqrt{\pi}\). So, the integral is equal to \(\frac{3}{4}\sqrt{\pi}\). (iii) \(\int_{0}^{\frac{1}{6}} \frac{\mathrm{dx}}{\mathrm{x} \ln ^{2} \mathrm{x}}\)

Use substitution method

Let \(u = \ln{x}\), so \(du = \frac{1}{x} dx\). Our integral becomes \(\int_{-\infty}^{-\ln{6}} \frac{1}{u^2} du\).

Evaluate the integral

Our integral is now \(\int_{-\infty}^{-\ln{6}} u^{-2} du = \left[-u^{-1}\right]_{-\infty}^{-\ln6} = -[-\ln{6}^{-1} - 0] = \frac{1}{\ln{6}}\). So, the integral is equal to \(\frac{1}{\ln{6}}\). (iv) \(\int_{-\infty}^{\infty} \frac{d x}{x^{2}+2 x+2}\)

Complete the square

Complete the square for the denominator: \(x^{2}+2x+2 = (x+1)^{2} + 1\)

Use trigonometric substitution

Use the trigonometric substitution: \(x+1 = \tan{u}\), so \(dx = (\sec{u})^2 du\). Our integral becomes \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(\sec{u})^2 du}{(\tan{u})^2 + 1}\).

Evaluate the integral

Our integral simplifies to \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sec{u})^2 du\). We can rewrite \((\sec{u})^2\) as \(1 + \tan^2{u}\). The integral becomes \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \tan^2{u}) du = \left[u + \tan{u}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\). So, the integral is equal to \(\pi\).