Tính các tích phân sau

a. ${\int }_0^2{\left(x^2+1\right)}^2\mathrm{d}x$

b. ${\int }_1^2\frac{2x^3+2x^2+1}{x+1}\mathrm{d}x$

c. ${\int }_0^{\pi }{\cos }^{2}x\mathrm{d}x$

d. ${\int }_0^1{2}^x\cdot 3^x\mathrm{d}x$

e. ${\int }_0^8\left|x^2-6x\right|\mathrm{d}x$

f. ${\int }_0^2\sqrt{e^{x+2}}\mathrm{d}x$

(AI giải đề cho tôi)
Câu 1 (Bài 12 – Tích phân), tính từng tích phân:

a. ${\int }_0^2(x^2+1{)}^{2}dx={\int }_0^2(x^4+2x^2+1)dx={\left[\frac{x^5}{5}+\frac{2x^3}{3}+x\right]}_0^2=\frac{32}{5}+\frac{16}{3}+2=\frac{96+80+30}{15}=\frac{206}{15}.$

b. Chia đa thức: $\frac{2x^3+2x^2+1}{x+1}=2x^2+\frac{1}{x+1}$. Do đó
${\int }_1^2\left(2x^2+\frac{1}{x+1}\right)dx={\left[\frac{2x^3}{3}+\ln (x+1)\right]}_1^2=\left(\frac{16}{3}+\ln 3\right)-\left(\frac{2}{3}+\ln 2\right)=\frac{14}{3}+\ln \frac{3}{2}.$

c. ${\int }_0^{\pi }{\cos }^{2}xdx={\int }_0^{\pi }\frac{1+\cos 2x}{2}dx={\left[\frac{x}{2}+\frac{\sin 2x}{4}\right]}_0^{\pi }=\frac{\pi }{2}.$

d. $2^x\cdot 3^x=6^x$, nên ${\int }_0^1{6}^{x}dx={\left[\frac{6^x}{\ln 6}\right]}_0^1=\frac{6-1}{\ln 6}=\frac{5}{\ln 6}.$

e. Trên $[0,6]$: $x^2-6x\le 0$; trên $[6,8]$: $x^2-6x\ge 0$.
${\int }_0^6(6x-x^2)dx={\left[3x^2-\frac{x^3}{3}\right]}_0^6=108-72=36$.
${\int }_6^8(x^2-6x)dx={\left[\frac{x^3}{3}-3x^2\right]}_6^8=\left(\frac{512}{3}-192\right)-(72-108)=-\frac{64}{3}+36=\frac{44}{3}$.
Tổng: $36+\frac{44}{3}=\frac{152}{3}.$

f. $\sqrt{e^{x+2}}=e^{(x+2)/2}$, nên ${\int }_0^2{e}^{(x+2)/2}dx={\left[2e^{(x+2)/2}\right]}_0^2=2e^2-2e=2e(e-1).$