Question

If the integral $\int \frac{dx}{\sqrt[4]{(x-1)^3(x+2)^5}}$ = $\frac{a}{9}(\frac{x-1}{x+2})^{1/4} + C$ then $a$ is equal (where $C$ is a constant of integration)

Answer 1

12

Answer 2

To solve the integral $\int \frac{dx}{\sqrt[4]{(x-1)^3(x+2)^5}}$, we first rewrite the expression using fractional exponents: $I = \int \frac{dx}{(x-1)^{3/4}(x+2)^{5/4}}$

This is a common type of integral where the substitution $t = \frac{x-1}{x+2}$ is effective. Let $t = \frac{x-1}{x+2}$. To find $dt$, we differentiate $t$ with respect to $x$: $\frac{dt}{dx} = \frac{(1)(x+2) - (x-1)(1)}{(x+2)^2} = \frac{x+2-x+1}{(x+2)^2} = \frac{3}{(x+2)^2}$. So, $dx = \frac{(x+2)^2}{3} dt$.

Now, we need to express the integrand in terms of $t$. The original denominator is $(x-1)^{3/4}(x+2)^{5/4}$. We can manipulate this to involve $\left(\frac{x-1}{x+2}\right)$: $(x-1)^{3/4}(x+2)^{5/4} = (x-1)^{3/4}(x+2)^{3/4}(x+2)^{2/4}$ $= \left(\frac{x-1}{x+2}\right)^{3/4}(x+2)^{3/4}(x+2)^{1/2}$ $= \left(\frac{x-1}{x+2}\right)^{3/4}(x+2)^{3/4+1/2}$ $= \left(\frac{x-1}{x+2}\right)^{3/4}(x+2)^{5/4}$. This is just the original denominator.

The key is to rearrange the integrand to get terms of $t$ and $dt$. We have $dx = \frac{(x+2)^2}{3} dt$. This means we need a term of $(x+2)^2$ in the denominator to replace it with $dt$. Let's rewrite the integral as: $I = \int \frac{1}{(x-1)^{3/4}(x+2)^{5/4}} dx$ Divide the numerator and denominator by $(x+2)^{3/4}$: $I = \int \frac{\frac{1}{(x+2)^{3/4}}}{\frac{(x-1)^{3/4}}{(x+2)^{3/4}}(x+2)^{5/4}} dx$ $I = \int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3/4} (x+2)^{5/4} (x+2)^{-3/4}} dx$ $I = \int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3/4} (x+2)^{5/4-3/4}} dx$ $I = \int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3/4} (x+2)^{2/4}} dx$ $I = \int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3/4} (x+2)^{1/2}} dx$

This manipulation is not leading to the required form for $dt$. Let's try a direct approach by factoring out $(x+2)$ from the denominator such that we can form $\left(\frac{x-1}{x+2}\right)$. $I = \int \frac{dx}{(x-1)^{3/4}(x+2)^{5/4}}$ Factor out $(x+2)^{3/4}$ from $(x+2)^{5/4}$ and combine it with $(x-1)^{3/4}$: $I = \int \frac{dx}{(x-1)^{3/4}(x+2)^{3/4}(x+2)^{2/4}}$ $I = \int \frac{dx}{\left(\frac{x-1}{x+2}\right)^{3/4}(x+2)^{3/4}(x+2)^{1/2}}$ $I = \int \frac{dx}{\left(\frac{x-1}{x+2}\right)^{3/4}(x+2)^{3/4+1/2}}$ $I = \int \frac{dx}{\left(\frac{x-1}{x+2}\right)^{3/4}(x+2)^{5/4}}$

This is also not helpful. The most straightforward way is to prepare the integrand for the substitution $t = \frac{x-1}{x+2}$ and $dx = \frac{(x+2)^2}{3} dt$. We need to get $\frac{1}{(x+2)^2}$ in the integrand. The denominator is $(x-1)^{3/4}(x+2)^{5/4}$. We can write $(x+2)^{5/4} = (x+2)^{8/4} (x+2)^{-3/4} = (x+2)^2 (x+2)^{-3/4}$. So, the integral becomes: $I = \int \frac{dx}{(x-1)^{3/4} (x+2)^2 (x+2)^{-3/4}}$ $I = \int \frac{dx}{(x+2)^2 \frac{(x-1)^{3/4}}{(x+2)^{3/4}}}$ $I = \int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3/4}} \frac{dx}{(x+2)^2}$

Now, substitute $t = \frac{x-1}{x+2}$ and $\frac{dx}{(x+2)^2} = \frac{1}{3} dt$: $I = \int \frac{1}{t^{3/4}} \cdot \frac{1}{3} dt$ $I = \frac{1}{3} \int t^{-3/4} dt$

Now, integrate $t^{-3/4}$: $I = \frac{1}{3} \left( \frac{t^{-3/4+1}}{-3/4+1} \right) + C$ $I = \frac{1}{3} \left( \frac{t^{1/4}}{1/4} \right) + C$ $I = \frac{1}{3} \cdot 4 t^{1/4} + C$ $I = \frac{4}{3} t^{1/4} + C$

Substitute back $t = \frac{x-1}{x+2}$: $I = \frac{4}{3} \left(\frac{x-1}{x+2}\right)^{1/4} + C$

The problem states that the integral is equal to $\frac{a}{9}(\frac{x-1}{x+2})^{1/4} + C$. Comparing our result with the given form: $\frac{4}{3} \left(\frac{x-1}{x+2}\right)^{1/4} + C = \frac{a}{9} \left(\frac{x-1}{x+2}\right)^{1/4} + C$ Equating the coefficients: $\frac{4}{3} = \frac{a}{9}$ $a = \frac{4}{3} \times 9$ $a = 4 \times 3$ $a = 12$

The final answer is $\boxed{12}$.

Explanation: The integral $\int \frac{dx}{\sqrt[4]{(x-1)^3(x+2)^5}}$ is transformed by factoring the denominator to create a term of the form $\left(\frac{x-1}{x+2}\right)$ and a term that can be absorbed into the differential $dx$. The integral is rewritten as $\int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3/4}} \frac{dx}{(x+2)^2}$. A substitution $t = \frac{x-1}{x+2}$ is made, leading to $\frac{dt}{dx} = \frac{3}{(x+2)^2}$, so $\frac{dx}{(x+2)^2} = \frac{1}{3} dt$. The integral simplifies to $\frac{1}{3} \int t^{-3/4} dt$. Integrating yields $\frac{1}{3} \frac{t^{1/4}}{1/4} + C = \frac{4}{3} t^{1/4} + C$. Substituting back $t = \frac{x-1}{x+2}$, we get $\frac{4}{3} \left(\frac{x-1}{x+2}\right)^{1/4} + C$. Comparing this with the given form $\frac{a}{9}(\frac{x-1}{x+2})^{1/4} + C$, we find $\frac{a}{9} = \frac{4}{3}$, which gives $a=12$.