Đáp án: $C$
Giải thích các bước giải:
Ta có:
$f(x)=\int f'(x)dx$
$\to f(x)=\int \dfrac{1}{(x^2+2x+3)\sqrt{x^2+2x+3}}dx$
$\to f(x)=\int \dfrac{1}{\left(x^2+2x+3\right)^{\dfrac{3}{2}}}dx$
$\to f(x)=\int \dfrac{1}{\left(\left(x+1\right)^2+2\right)^{\dfrac{3}{2}}}dx$
Đặt $x+1=\sqrt{2}\tan(v)\to d(x)=\sqrt{2}\dfrac{1}{\cos^2v}dv$
$\to \sin v=\dfrac{x+1}{\sqrt{x^2+2x+3}}$
$\to f(x)=\int \dfrac{1}{\left(\left(\sqrt{2}\tan(v)\right)^2+2\right)^{\dfrac{3}{2}}}\cdot\sqrt{2}\dfrac{1}{\cos^2v}dv$
$\to f(x)=\int \dfrac{1}{\left(2\tan^2(v)+2\right)^{\dfrac{3}{2}}}\cdot\sqrt{2}\dfrac{1}{\cos^2v}dv$
$\to f(x)=\int \dfrac{1}{\left(2(\tan^2(v)+1)\right)^{\dfrac{3}{2}}}\cdot\sqrt{2}\dfrac{1}{\cos^2v}dv$
$\to f(x)=\int \dfrac{1}{\left(2\cdot\dfrac{1}{\cos^2(v)}\right)^{\dfrac{3}{2}}}\cdot\sqrt{2}\dfrac{1}{\cos^2v}dv$
$\to f(x)=\int \dfrac{\cos \left(v\right)}{2}dv$
$\to f(x)=\dfrac{1}{2}\sin \left(v\right)+C$
$\to f(x)=\dfrac{x+1}{2\sqrt{x^2+2x+3}}+C$
Mà $f(-1)=2$
$\to f(-1)=\dfrac{-1+1}{2\sqrt{(-1)^2+2\cdot (-1)+3}}+C=2\to C=2$
$\to f(x)=\dfrac{x+1}{2\sqrt{x^2+2x+3}}+2$
$\to \int^5_3f(x)dx=\int^5_3\dfrac{x+1}{2\sqrt{x^2+2x+3}}+2dx$
$\to \int^5_3f(x)dx=\int^5_3\dfrac12\cdot \dfrac{2x+2}{2\sqrt{x^2+2x+3}}+2dx$
$\to \int^5_3f(x)dx=\dfrac12\sqrt{x^2+2x+3}+2x\Bigg|^5_3$
$\to \int^5_3f(x)dx=\dfrac{\sqrt{38}-3\sqrt{2}}{2}+4$
$\to \int^5_3f(x)dx=\dfrac{\sqrt{38}-3\sqrt{2}+8}{2}$
$\to \int^5_3f(x)dx=\dfrac{\sqrt{38}-\sqrt{18}+8}{2}$
$\to a=38, b=18, c=8$
$\to a+b+c=38+18+8=64$
$\to C$
