Question

Evaluate the following integrals: (i) \(\int \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+1\right) \sqrt{\mathrm{x}}}\) (ii) \(\int \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+5 \mathrm{x}+6\right) \sqrt{\mathrm{x}+1}}\) (iii) \(\int \frac{d x}{\left(x^{2}-4\right) \sqrt{x+1}}\)

Short answer

Question: Evaluate each of the following integrals. (i) $\int \frac{dx}{(x^2+1) \sqrt{x}}$ (ii) $\int \frac{dx}{(x^2+5x+6)\sqrt{x+1}}$ (iii) $\int \frac{dx}{(x^2 - 4)\sqrt{x+1}}$ Answer: (i) $2 \arctan(\sqrt{x^2}) + C$ (ii) $2\ln(\sqrt{x+1}+\sqrt{x+2}) - \ln(\sqrt{x+1}+\sqrt{x+3}) + C$ (iii) $\frac{1}{2}\ln \left(\frac{(x+2)+\sqrt{(x+1)(x+2)}}{(x-2)+\sqrt{(x+1)(x-2)}}\right) + C$

Step-by-step solution

Integral (i)

To solve this integral, we'll use the substitution method. Let's use the variable \(u\) for the substitution. Let \(u = \sqrt{x}\). Thus, \(u^2 = x\). Now differentiate both sides of the equation to find \(du\): \(2u\, du = dx\). Now, we can replace each variable in the original integral: $$ \int \frac{dx}{(x^2+1) \sqrt{x}} = \int \frac{2u\, du}{(u^4+1)u}. $$ We can simplify the expression and solve the integral: $$ \int \frac{2u\, du}{u^5+u} = 2\int \frac{du}{u^4+1}. $$ Now let us use another substitution to make it easier to solve the integral. Let \(v = u^2\). Then, \(2u\, du = dv\). So, the integral becomes $$ 2\int \frac{dv}{v^2+1} = 2\arctan{v} + C = 2\arctan{(\sqrt{x^2})+C}, $$ where C is the constant of integration. This is the final answer.

Integral (ii)

To solve this integral, we first need to factor the quadratic expression in the denominator. We'll have: $$ x^2+5x+6 = (x+3)(x+2). $$ Now, we use partial fractions decomposition to rewrite the fraction in the integral as the sum of two separate fractions: $$ \frac{1}{(x^2+5x+6)\sqrt{x+1}} = \frac{A}{(x+2)\sqrt{x+1}} + \frac{B}{(x+3)\sqrt{x+1}}, $$ where A and B are constants to be determined. Now, multiplying both sides by \((x^2+5x+6)\sqrt{x+1}\), we have: $$ 1 = A(x+3)(x+1) + B(x+2)(x+1). $$ By comparing the coefficients of the polynomial, we find that \(A = 1\) and \(B = -1\). So, the integral becomes: $$ \int \frac{dx}{(x^2+5x+6)\sqrt{x+1}} = \int \frac{dx}{(x+2)\sqrt{x+1}} - \int \frac{dx}{(x+3)\sqrt{x+1}}. $$ Now, to solve these integrals, we use substitution. Let \(u = \sqrt{x+1}\), then \(u^2 = x+1\), and \(2u\, du = dx\). We can rewrite the two integrals as: $$ \int \frac{dx}{(x+2)\sqrt{x+1}} = 2\int \frac{u\, du}{(u^2+1)}, $$ and $$ \int \frac{dx}{(x+3)\sqrt{x+1}} = 2\int \frac{u\, du}{(u^2+2)}. $$ Now we can solve both integrals: $$ 2\int \frac{u\, du}{(u^2+1)} = 2\ln(u+\sqrt{u^2+1}) + C_1, $$ and $$ 2\int \frac{u\, du}{(u^2+2)} = \ln(u+\sqrt{u^2+2}) + C_2, $$ where \(C_1\) and \(C_2\) are constants of integration. Now substituting back \(u = \sqrt{x+1}\), we get: $$ 2\ln(\sqrt{x+1}+\sqrt{x+2}) - \ln(\sqrt{x+1}+\sqrt{x+3}) + C. $$

Integral (iii)

To solve this integral, first rewrite the denominator as a difference of squares: $$ x^2 - 4 = (x+2)(x-2). $$ Then, use partial fractions to rewrite the expression as the sum of two separate fractions: $$ \frac{1}{(x^2 - 4)\sqrt{x+1}} = \frac{A}{(x+2)\sqrt{x+1}} + \frac{B}{(x-2)\sqrt{x+1}}. $$ After solving for A and B using a similar method as in the second integral, we find that \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\). Now, we need to solve: $$ \int \frac{dx}{(x^2-4)\sqrt{x+1}} = \frac{1}{2} \int \frac{dx}{(x+2)\sqrt{x+1}} - \frac{1}{2}\int \frac{dx}{(x-2)\sqrt{x+1}}. $$ Now, just like we did above, we can use the substitution method to evaluate these integrals, and the solution will be: $$ \frac{1}{2}\ln \left(\frac{(x+2)+\sqrt{(x+1)(x+2)}}{(x-2)+\sqrt{(x+1)(x-2)}}\right) + C, $$ where C is the constant of integration.