# Question 45e53

Nov 19, 2017

$\int \frac{1}{{\left(x - 1\right)}^{3} \cdot {\left(x + 2\right)}^{5}} ^ \left(\frac{1}{4}\right) = \frac{4}{3} \cdot {\left(\frac{x - 1}{x + 2}\right)}^{\frac{1}{4}} + C$

#### Explanation:

$\int \frac{1}{{\left(x - 1\right)}^{3} \cdot {\left(x + 2\right)}^{5}} ^ \left(\frac{1}{4}\right)$

=$\int \frac{\mathrm{dx}}{{\left(x - 1\right)}^{\frac{3}{4}} \cdot {\left(x + 2\right)}^{\frac{5}{4}}}$

After using $u = {\left(\frac{x - 1}{x + 2}\right)}^{\frac{1}{4}}$ substitution,

${u}^{4} = \frac{x - 1}{x + 2}$

${u}^{4} \cdot \left(x + 2\right) = x - 1$

${u}^{4} \cdot x + 2 {u}^{4} = x - 1$

$2 {u}^{4} + 1 = x - {u}^{4} \cdot x$

$x = \frac{2 {u}^{4} + 1}{1 - {u}^{4}}$ and,

$\mathrm{dx} = \frac{8 {u}^{3} \cdot \left(1 - {u}^{4}\right) - \left(- 4 {u}^{3}\right) \cdot \left(2 {u}^{4} + 1\right)}{1 - {u}^{4}} ^ 2 \cdot \mathrm{du}$

=$\frac{12 {u}^{3} \cdot \mathrm{du}}{1 - {u}^{4}} ^ 2$

Also denominator became,

${\left(x - 1\right)}^{\frac{3}{4}} \cdot {\left(x + 2\right)}^{\frac{5}{4}}$

=${\left(\frac{2 {u}^{4} + 1}{1 - {u}^{4}} - 1\right)}^{\frac{3}{4}} \cdot {\left(\frac{2 {u}^{4} + 1}{1 - {u}^{4}} + 2\right)}^{\frac{5}{4}}$

=$\frac{9 {u}^{3}}{1 - {u}^{4}} ^ 2$

Thus,

$\int \frac{\mathrm{dx}}{{\left(x - 1\right)}^{\frac{3}{4}} \cdot {\left(x + 2\right)}^{\frac{5}{4}}}$

$\int \frac{\frac{12 {u}^{3} \cdot \mathrm{du}}{1 - {u}^{4}} ^ 2}{\frac{9 {u}^{3}}{1 - {u}^{4}} ^ 2}$

=$\int \frac{4}{3} \cdot \mathrm{du}$

=$\frac{4}{3} \cdot u + C$

=$\frac{4}{3} \cdot {\left(\frac{x - 1}{x + 2}\right)}^{\frac{1}{4}} + C$

Dec 5, 2017

$\frac{4}{3} {\left(\frac{x - 1}{x - 2}\right)}^{\frac{1}{4}} + C , \mathmr{and} , \frac{4}{3} \sqrt[4]{\frac{x - 1}{x - 2}} + C .$

#### Explanation:

Here is another Method to find the Integral.

Let, $I = \int \frac{1}{{\left(x - 1\right)}^{3} {\left(x + 2\right)}^{5}} ^ \left(\frac{1}{4}\right) \mathrm{dx} .$

$\therefore I = \int \frac{1}{\left\{{\left(x - 1\right)}^{4} / \left(x - 1\right)\right\} \left\{{\left(x + 2\right)}^{4} \left(x + 2\right)\right\}} ^ \left(\frac{1}{4}\right) \mathrm{dx} ,$

$= \int \left[\left\{\frac{1}{{\left(x - 1\right)}^{4} {\left(x + 2\right)}^{4}} ^ \left(\frac{1}{4}\right)\right\} {\left\{\frac{x - 1}{x + 2}\right\}}^{\frac{1}{4}}\right] \mathrm{dx} ,$

$= \int \left\{\frac{1}{\left(x - 1\right) \left(x + 2\right)} \cdot {\left(\frac{x - 1}{x + 2}\right)}^{\frac{1}{4}}\right\} \mathrm{dx} .$

Now, we substitute $\frac{x - 1}{x + 2} = {t}^{4.}$

$\therefore d \left\{\frac{x - 1}{x + 2}\right\} = d \left({t}^{4}\right) , i . e . ,$

$\left[\frac{\left(x + 2\right) \cdot 1 - \left(x - 1\right) \cdot 1}{x + 2} ^ 2\right] \mathrm{dx} = 4 {t}^{3} \mathrm{dt} , \mathmr{and} ,$

$\frac{3}{x + 2} ^ 2 \mathrm{dx} = 4 {t}^{3} \mathrm{dt} \Rightarrow \mathrm{dx} = \frac{4}{3} {t}^{3} {\left(x + 2\right)}^{2} \mathrm{dt} .$

Also, $\frac{x - 1}{x + 2} = {t}^{4} \Rightarrow \left(x - 1\right) = {t}^{4} \left(x + 2\right) .$

$\therefore I = \int \left\{\frac{1}{{t}^{4} {\left(x + 2\right)}^{2}} \cdot {\left({t}^{4}\right)}^{\frac{1}{4}}\right\} \frac{4}{3} \cdot {t}^{3} {\left(x + 2\right)}^{2} \mathrm{dt} ,$

$= \frac{4}{3} \int 1 \mathrm{dt} ,$

$= \frac{4}{3} t .$

$\Rightarrow I = \frac{4}{3} {\left(\frac{x - 1}{x - 2}\right)}^{\frac{1}{4}} + C , \mathmr{and} , \frac{4}{3} \sqrt[4]{\frac{x - 1}{x - 2}} + C .$

Enjoy Maths.!