Integrate number 1

Let the integral be $I$ . Then:

$\begin{aligned} I & = \int_0^1 \color{#3D99F6}{5^x}(\color{#D61F06}{3x^2+5x+4}) \color{#3D99F6}{\space dx} \quad \quad \small \color{#3D99F6}{\text{By integration by parts: }} \color{#D61F06}{u = 3x^2+5x+4} \color{#3D99F6}{, \space dv = 5^x \space dx} \\ & = \left[ (3x^2+5x+4)\frac{5^x}{\ln 5}\right]_0^1 - \frac{1}{\ln 5} \int_0^1 \color{#3D99F6}{5^x}(\color{#D61F06}{6x+5}) \color{#3D99F6}{\space dx} \quad \quad \small \color{#D61F06}{u = 6x+5} \color{#3D99F6}{, \space dv = 5^x \space dx} \\ & = \frac{56}{\ln 5} - \frac{1}{\ln 5} \left[ (6x+5)\frac{5^x}{\ln 5}\right]_0^1 + \frac{6}{\ln^2 5} \int_0^1 5^x \space dx \\ & = \frac{56}{\ln 5} - \frac{50}{\ln^2 5} + \frac{6}{\ln^2 5} \left[\frac{5^x}{\ln 5}\right]_0^1 \\ & = \frac{56}{\ln 5} - \frac{50}{\ln^2 5} + \frac{24}{\ln^3 5} \end{aligned}$

$\Rightarrow A+B+C = 24-50+56 = \boxed{30}$

$\displaystyle \int_{0}^{1} 5^{x}(3x^2+5x+4)dx$ $=\displaystyle \int_{0}^{1} (e^{x\ln(5)}\cdot 3x^2 + e^{x\ln(5)}\cdot 5x + 4e^{x\ln(5)})dx$ $=(3\frac{x^2e^{x\ln(5)}}{\ln(5)}|_{0}^{1} +(-\frac{6}{\ln(5)}+5)\displaystyle \int_{0}^{1} xe^{x\ln(5)}dx+\frac{4}{\ln(5)}(5-1)$ $=\frac{31}{\ln(5)}+(-\frac{6}{\ln(5)}+5)((\frac{xe^{x\ln(5)}}{\ln(5)}|_{0}^{1}-\frac{1}{\ln(5)}\displaystyle \int_{0}^{1} e^{x\ln(5)}dx)$ $=\frac{31}{\ln(5)}+(-\frac{6}{\ln(5)}+5)(\frac{5}{\ln(5)}-\frac{4}{(\ln(5))^2})$ $=\frac{24}{(\ln(5))^3}+\frac{-50}{(\ln(5))^2}+\frac{56}{\ln(5)}$ So we have: $A=24$ , $B=-50$ and $C=56$ which gives: $A+B+C=24-50+56=30$

A=24,B=-50 and C=56. answer=24-50+56=30.