The integral e^3_e(2x)+5e^2_e(2x)e^4_ex+5e^3_ex-7e^2_ex,dx, x0, is equal to (where c is a…

Question

The integral $\displaystyle\int\frac{e^{3\log_e(2x)}+5e^{2\log_e(2x)}}{e^{4\log_ex}+5e^{3\log_ex}-7e^{2\log_ex}}\,dx$, $x>0$, is equal to (where $c$ is a constant of integration):

Accepted Answer

Using $e^{a\log_ex}=x^a$:

$$I = \int\frac{(2x)^3+5(2x)^2}{x^4+5x^3-7x^2}\,dx = \int\frac{8x^3+20x^2}{x^4+5x^3-7x^2}\,dx = \int\frac{4x^2(2x+5)}{x^2(x^2+5x-7)}\,dx = \int\frac{4(2x+5)}{x^2+5x-7}\,dx$$

Let $t=x^2+5x-7$, $dt=(2x+5)\,dx$:

$$I = \int\frac{4\,dt}t = 4\log_e|t|+c = 4\log_e|x^2+5x-7|+c$$

Solutions · 1