Let I(x) = dx(x-11)^11/13(x+15)^15/13.

Question

Let $I(x) = \displaystyle\int\frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}$. If $I(37)-I(24) = \dfrac14\left(\dfrac1{b^{1/13}}-\dfrac1{c^{1/13}}\right)$, $b,c\in\mathbb{N}$, then $3(b+c)$ is equal to:

Accepted Answer

Substitute $\dfrac{x-11}{x+15}=t$, giving $\dfrac{26}{(x+15)^2}dx=dt$, and

$$I(x) = \frac1{26}\int t^{-11/13}\,dt = \frac14\left(\frac{x-11}{x+15}\right)^{2/13}+C$$

$$I(37)-I(24) = \frac14\left[\left(\frac{26}{52}\right)^{2/13}-\left(\frac{13}{39}\right)^{2/13}\right] = \frac14\left[\frac1{4^{1/13}}-\frac1{9^{1/13}}\right]$$

So $b=4,\,c=9$, and $3(b+c)=39$.

Solutions · 1