Could you help me solve?

Answer:
1.14200738982×10^26
Step-by-step explanation:
Substitution can make this integral much easier to evaluate.
Let u = 7x² -x. Then du = (14x -1)dx. The limits on x become different limits for u:
for x = 1: u = 7(1²) -1 = 6
for x = 3: u = 7(3²) -3 = 60
[tex]\displaystyle\int_1^3{(14x-1)e^{(7x^2-x)}}\,dx=\int_6^{60}{e^u}\,du=\left.e^u\right|^{60}_6\\\\=e^{60}-e^6\approx e^{60}\approx\boxed{1.14200738982\times10^{26}}[/tex]