1. **State the problem:** We need to evaluate the definite integral $$L=\int_0^{1.3} \left(1+0.093636x\right)^{-0.98} \, dx.$$\n\n2. **Recall the formula:** For an integral of the form $$\int (a+bx)^n \, dx,$$ the antiderivative is $$\frac{(a+bx)^{n+1}}{b(n+1)} + C,$$ provided $$n \neq -1.$$\n\n3. **Apply the formula:** Here, $$a=1,$$ $$b=0.093636,$$ and $$n=-0.98.$$ Since $$n+1 = 0.02 \neq 0,$$ the formula applies.\n\n4. **Find the antiderivative:** $$F(x) = \frac{(1+0.093636x)^{0.02}}{0.093636 \times 0.02} = \frac{(1+0.093636x)^{0.02}}{0.00187272}.$$\n\n5. **Evaluate the definite integral:** $$L = F(1.3) - F(0) = \frac{(1+0.093636 \times 1.3)^{0.02} - (1)^{0.02}}{0.00187272}.$$\nCalculate inside the parentheses:\n$$1 + 0.093636 \times 1.3 = 1 + 0.1217268 = 1.1217268.$$\n\n6. **Calculate powers:**\n$$1.1217268^{0.02} = e^{0.02 \ln(1.1217268)}.$$\nCalculate $$\ln(1.1217268) \approx 0.115.$$ So, $$1.1217268^{0.02} \approx e^{0.02 \times 0.115} = e^{0.0023} \approx 1.0023.$$\n\n7. **Substitute back:**\n$$L = \frac{1.0023 - 1}{0.00187272} = \frac{0.0023}{0.00187272} \approx 1.228.$$\n\n**Final answer:** $$L \approx 1.228.$$