Question

If $\int \frac{dx}{\sqrt[6]{(x+1)^5(x-3)^7}}=a(\frac{x+1}{x-3})^b+C$, where $C$ is constant of integration, then the value of $3b-a$ is

Answer 1

2

Answer 2

The given integral is $I = \int \frac{dx}{\sqrt[6]{(x+1)^5(x-3)^7}}$. We can rewrite the integral using fractional exponents: $I = \int (x+1)^{-5/6} (x-3)^{-7/6} dx$

The problem asks us to express the integral in the form $a(\frac{x+1}{x-3})^b+C$. This suggests a substitution involving the ratio $\frac{x+1}{x-3}$.

Let $t = \frac{x+1}{x-3}$. To find $dt$, we differentiate $t$ with respect to $x$ using the quotient rule: $dt = \frac{d}{dx}\left(\frac{x+1}{x-3}\right) dx = \frac{(x-3)(1) - (x+1)(1)}{(x-3)^2} dx$ $dt = \frac{x-3-x-1}{(x-3)^2} dx = \frac{-4}{(x-3)^2} dx$ From this, we can express $dx$ in terms of $dt$: $dx = -\frac{1}{4}(x-3)^2 dt$.

Now, we need to express the integrand $(x+1)^{-5/6} (x-3)^{-7/6}$ in terms of $t$. From the substitution $t = \frac{x+1}{x-3}$, we have $x+1 = t(x-3)$. Substitute this into the integrand: $(x+1)^{-5/6} (x-3)^{-7/6} = (t(x-3))^{-5/6} (x-3)^{-7/6}$ $= t^{-5/6} (x-3)^{-5/6} (x-3)^{-7/6}$ $= t^{-5/6} (x-3)^{-(5/6 + 7/6)}$ $= t^{-5/6} (x-3)^{-12/6}$ $= t^{-5/6} (x-3)^{-2}$ $= \frac{t^{-5/6}}{(x-3)^2}$

Now substitute this back into the integral $I$: $I = \int \frac{t^{-5/6}}{(x-3)^2} \left(-\frac{1}{4}(x-3)^2 dt\right)$ The term $(x-3)^2$ cancels out: $I = \int -\frac{1}{4} t^{-5/6} dt$ $I = -\frac{1}{4} \int t^{-5/6} dt$

Now, integrate $t^{-5/6}$ with respect to $t$: $\int t^{-5/6} dt = \frac{t^{-5/6+1}}{-5/6+1} + C = \frac{t^{1/6}}{1/6} + C = 6t^{1/6} + C$.

Substitute this back into the expression for $I$: $I = -\frac{1}{4} (6t^{1/6}) + C$ $I = -\frac{6}{4} t^{1/6} + C$ $I = -\frac{3}{2} t^{1/6} + C$

Finally, substitute back $t = \frac{x+1}{x-3}$: $I = -\frac{3}{2} \left(\frac{x+1}{x-3}\right)^{1/6} + C$.

This result is in the form $a(\frac{x+1}{x-3})^b+C$. By comparing, we find: $a = -\frac{3}{2}$ $b = \frac{1}{6}$

We need to find the value of $3b-a$: $3b-a = 3\left(\frac{1}{6}\right) - \left(-\frac{3}{2}\right)$ $3b-a = \frac{3}{6} + \frac{3}{2}$ $3b-a = \frac{1}{2} + \frac{3}{2}$ $3b-a = \frac{1+3}{2}$ $3b-a = \frac{4}{2}$ $3b-a = 2$.

The final answer is $\boxed{2}$.

Explanation of the solution: The integral is of the form $\int (x+1)^m (x-3)^n dx$. Given the desired form of the result, we use the substitution $t = \frac{x+1}{x-3}$.

1. Calculate $dt$ in terms of $dx$ and $(x-3)^2$: $dt = \frac{-4}{(x-3)^2} dx$.
2. Express the integrand in terms of $t$: $(x+1)^{-5/6}(x-3)^{-7/6} = t^{-5/6}(x-3)^{-2}$.
3. Substitute $t$ and $dt$ into the integral: $\int \frac{t^{-5/6}}{(x-3)^2} \left(-\frac{1}{4}(x-3)^2 dt\right) = -\frac{1}{4} \int t^{-5/6} dt$.
4. Integrate: $-\frac{1}{4} \left(\frac{t^{1/6}}{1/6}\right) + C = -\frac{3}{2} t^{1/6} + C$.
5. Substitute back $t = \frac{x+1}{x-3}$: $-\frac{3}{2} \left(\frac{x+1}{x-3}\right)^{1/6} + C$.
6. Compare with $a(\frac{x+1}{x-3})^b+C$ to find $a = -\frac{3}{2}$ and $b = \frac{1}{6}$.
7. Calculate $3b-a = 3(\frac{1}{6}) - (-\frac{3}{2}) = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$.