1. The problem is to evaluate the integral $$\int \frac{6x}{e^{2x}+1} \, dx.$$\n\n2. We will use substitution to solve this integral. Let $$u = e^{2x} + 1.$$ Then, differentiate $$u$$ with respect to $$x$$:\n$$\frac{du}{dx} = 2e^{2x} \implies du = 2e^{2x} dx.$$\n\n3. We need to express $$6x dx$$ in terms of $$du$$ and $$u$$. Notice that $$6x$$ is not directly related to $$du$$, so we try integration by parts instead.\n\n4. Let $$I = \int \frac{6x}{e^{2x}+1} dx.$$ Set:\n- $$u = 6x$$ so $$du = 6 dx,$$\n- $$dv = \frac{1}{e^{2x}+1} dx,$$ and we need to find $$v = \int \frac{1}{e^{2x}+1} dx.$$\n\n5. To find $$v$$, substitute $$t = e^{2x}$$, so $$dt = 2e^{2x} dx = 2t dx \implies dx = \frac{dt}{2t}.$$\nThen,\n$$v = \int \frac{1}{t+1} \cdot \frac{dt}{2t} = \frac{1}{2} \int \frac{1}{t(t+1)} dt.$$\n\n6. Use partial fractions:\n$$\frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1}.$$\nMultiply both sides by $$t(t+1)$$:\n$$1 = A(t+1) + Bt = (A+B)t + A.$$\nEquate coefficients:\n$$A = 1,$$\n$$A + B = 0 \implies B = -1.$$\n\n7. So,\n$$v = \frac{1}{2} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt = \frac{1}{2} (\ln|t| - \ln|t+1|) + C = \frac{1}{2} \ln \left| \frac{t}{t+1} \right| + C.$$\nSubstitute back $$t = e^{2x}$$:\n$$v = \frac{1}{2} \ln \left| \frac{e^{2x}}{e^{2x}+1} \right| + C.$$\n\n8. Now apply integration by parts formula:\n$$I = uv - \int v du = 6x \cdot \frac{1}{2} \ln \left| \frac{e^{2x}}{e^{2x}+1} \right| - \int \frac{1}{2} \ln \left| \frac{e^{2x}}{e^{2x}+1} \right| \cdot 6 dx.$$\nSimplify:\n$$I = 3x \ln \left| \frac{e^{2x}}{e^{2x}+1} \right| - 3 \int \ln \left| \frac{e^{2x}}{e^{2x}+1} \right| dx.$$\n\n9. The remaining integral is complicated, so the integral in closed form involves this expression.\n\nFinal answer:\n$$\int \frac{6x}{e^{2x}+1} dx = 3x \ln \left| \frac{e^{2x}}{e^{2x}+1} \right| - 3 \int \ln \left| \frac{e^{2x}}{e^{2x}+1} \right| dx + C.$$