Question

integrala din(4x-3)×e^x ×dx =?
integrala din(x^2+1)×e^x×dx=?​

Answer 1

$\displaystyle \boxed{\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx}\\\\ a)~\int (4x-3)e^xdx\\\\ f(x)=4x-3 \Rightarrow f'(x)=(4x-3)' =4 \cdot x' -3'=4 \cdot 1-0=4\\\\ g'(x)=e^x \Rightarrow g(x)=\int e^x dx=e^x+C\\\\ \int(4x-3)e^xdx=(4x-3)e^x-\int 4e^xdx=(4x-3)e^x-4\int e^xdx=\\\\=4xe^x-3e^x-4e^x=4xe^x-7e^x+C$

$\displaystyle b)~\int \Big(x^2+1\Big)e^x dx \\\\ f(x)=x^2+1 \Rightarrow f'(x)=\Big(x^2+1\Big)'=\Big(x^2\Big)' +1'=2x+0=2x\\\\ g'(x)=e^x \Rightarrow g(x)=\int e^xdx=e^x+C\\\\ \int\Big(x^2+1\Big)e^x=\Big(x^2+1\Big)e^x-\int 2xe^x=\Big(x^2+1\Big)e^x-2\int xe^x$

$\displaystyle \int xe^x\\\\ f(x)=x \Rightarrow f'(x)=x'=1\\\\ g'(x)=e^x\Rightarrow g(x)=\int e^xdx=e^x+C\\\\ \int\Big(x^2+1\Big)e^x=\Big(x^2+1\Big)e^x-\int 2xe^x=\Big(x^2+1\Big)e^x-2\int xe^x=\\\\=x^2e^x+e^x-2\Bigg(xe^x-\int e^x\Bigg)=x^2e^x+e^x-2\Big(xe^x-e^x\Big)=\\\\=x^2e^x+e^x-2xe^x+2e^x=x^2e^x-2xe^x+3e^x+C$