Question

Evaluate the following integrals : (i) \(\int \frac{x^{2}}{\sqrt{1-x}} d x\) (ii) \(\int \sqrt{\frac{x}{1-x^{3}}} d x\) (iii) \(\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x\) (iv) \(\int\left(27 e^{9 x}+e^{12 x}\right)^{1 / 3} d x\)

Short answer

The solutions for the given integrals are: (i) \(\int \frac{x^{2}}{\sqrt{1-x}} d x = -2(1-x)^{1/2}-\frac{4}{3} (1-x)^{3/2}+\frac{2}{5} (1-x)^{5/2}+C\) (ii) \(\int \sqrt{\frac{x}{1-x^{3}}} d x = -2\Gamma\left(\frac{5}{6}\right)\) (iii) \(\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x = x-2\ln|x+1|-(x+1)^{-1}+C\) (iv) \(\int\left(27 e^{9 x}+e^{12 x}\right)^{1 / 3} d x = \frac{3}{4}\left(27 e^{9 x}+e^{12 x}\right)^{4 / 3}+C\)

Step-by-step solution

Substitution

Let \(u = 1-x\), so \(x = 1-u\) and \(d x = -d u\). Now we can rewrite the integral in terms of \(u\):$$\int \frac{(1-u)^{2}}{\sqrt{u}} (-d u)$$

Simplify and Integrate

Now, we simplify the integral and integrate with respect to \(u\):$$-\int \frac{(1-2u+u^2)}{\sqrt{u}} d u=-\int (u^{-1/2}-2u^{1/2}+u^{3/2}) d u$$Now integrate each term:$$-\left[\int u^{-1/2} d u - 2 \int u^{1/2} d u + \int u^{3/2} d u \right]$$After integrating, we get:$$-2u^{1/2}-\frac{4}{3} u^{3/2}+\frac{2}{5} u^{5/2}+C$$

Reverse Substitution

Now we substitute \(u=1-x\) back:$$-2(1-x)^{1/2}-\frac{4}{3} (1-x)^{3/2}+\frac{2}{5} (1-x)^{5/2}+C$$So:$$\int \frac{x^{2}}{\sqrt{1-x}} d x = -2(1-x)^{1/2}-\frac{4}{3} (1-x)^{3/2}+\frac{2}{5} (1-x)^{5/2}+C$$ (ii) \(\int \sqrt{\frac{x}{1-x^{3}}} d x\)

Substitution

Let \(u = x^3\), so \(x = u^{1/3}\) and \(d x = \frac{1}{3}u^{-2/3} d u\). Now rewrite the integral in terms of \(u\):$$\int \sqrt{\frac{u^{1/3}}{1-u}} \frac{1}{3}u^{-2/3} d u$$

Simplify and Integrate

Now, we simplify the integral and integrate with respect to \(u\):$$\frac{1}{3}\int u^{-1/6}(1-u)^{-1/2} d u$$We can now integrate this expression. Let's use the substitution \(v = 1-u\), so \(d v = -d u\). This new integral becomes$$-\frac{1}{3}\int v^{1/2}(1-v)^{-1/6} d v = -\frac{1}{3}\int v^{\frac{1}{2}-1} (1-v)^{-\frac{1}{6}} d v$$This is in the format of the beta function, so integrating:$-\frac{1}{3} B\left(\frac{1}{2},\frac{5}{6}\right) = -\frac{1}{3} \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{5}{6}\right)}{\Gamma\left(\frac{3}{2}\right)}=-\frac{1}{3} \frac{\sqrt{\pi} \Gamma\left(\frac{5}{6}\right)}{\frac{1}{2}\sqrt\pi}$$Hence, the integral is equal to$$-2\Gamma\left(\frac{5}{6}\right)$$ (iii) \(\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x\)

Simplify and Integrate

In this integral, we can directly work on the polynomial inside the integral. First, we expand the numerator:$$\int \frac{x^2+3x+1}{(x+1)^2} d x =\int \frac{(x+1)^2-2(x+1)+1}{(x+1)^2} d x$$Now, we can simplify the integral:$$\int 1- 2\frac{x+1}{(x+1)^{2}}+\frac{1}{(x+1)^{2}} d x$$Integrating each term, we get:$$x-2\int \frac{x+1}{(x+1)^{2}} d x + \int \frac{1}{(x+1)^{2}} d x$$Integrating each term, we get:$$x-2\int \frac{1}{x+1} d x+\int (x+1)^{-2} d x$$So, the integral becomes:$$x-2\ln|x+1|-(x+1)^{-1}+C$$ (iv) \(\int\left(27 e^{9 x}+e^{12 x}\right)^{1 / 3} d x\)

Substitution

Let \(u=9x\), so \(d x=1/9 \ d u\). The integral becomes:$$\frac{1}{9}\int\left(27 e^u+e^{4 u/3}\right)^{1/3} d u$$

Simplify and Integrate

Now let \(v=e^{u/3}\), so \(d v=\frac 13 e^{u/3} d u=d u\). Then the integral changes to$$\int\left(27 v^3+v^{4}\right)^{1/3} d v$$Now let \(w=27 v^3+v^4\), then \(d w= (81 v^2+4 v^3) d v\). The integral becomes:$$\int w^{1/3} d w$$

Integrate and Reverse Substitution

Easily integrate this integral,$$\frac{3}{4}w^{4/3}+C$$Then reverse the substitutions,$$\frac{3}{4}\left(27 e^{u}+e^{4 u/3}\right)^{4 / 3}+C$$$$=\frac{3}{4}\left(27 e^{9x}+e^{12x}\right)^{4 / 3}+C$$ So the integral is equal to:$$\frac{3}{4}\left(27 e^{9 x}+e^{12 x}\right)^{4 / 3}+C$$