Question 13(d)Question 13(d) Using the substitution u=ex+2e–xu = e^x + 2e^{–x}u=ex+2e–x, and considering u2u^2u2, find ∫e3x−2ex4+8e2x+e4xdx\int \frac{e^{3x} - 2e^x}{4 + 8e^{2x} + e^{4x}} dx∫4+8e2x+e4xe3x−2exdx Solution We note that dudx=ex−2e−x\frac{du}{dx} = e^x-2e^{-x}dxdu=ex−2e−x and u2=(ex)2+2(ex)(2e−x)+(2e−x)2u2=e2x+4+4e−2x\begin{aligned} u^2 &= \left(e^x\right)^2 + 2\left(e^x\right)\left(2e^{-x}\right) + \left(2e^{-x}\right)^2 \\ u^2 &= e^{2x} + 4 + 4e^{-2x} \end{aligned}u2u2=(ex)2+2(ex)(2e−x)+(2e−x)2=e2x+4+4e−2x ... Log in or sign up to see more
