Giải thích các bước giải:
Câu 1:
Ta có:
$\displaystyle\int f(x)dx=\displaystyle\int\dfrac{2x}{x^2+1}dx$
$\to \displaystyle\int f(x)dx=\displaystyle\int\dfrac{1}{x^2+1}d(x^2)$
$\to \displaystyle\int f(x)dx=\ln(x^2+1)+C$
$\to D$
Câu 2:
Ta có:
$\displaystyle\int f(x)dx=\displaystyle\int\dfrac{2x+1}{x^2+x+4}dx$
$\to \displaystyle\int f(x)dx=\displaystyle\int\dfrac{1}{x^2+x+4}d(x^2+x+4)$
$\to \displaystyle\int f(x)dx=\ln(x^2+x+4)+C$
$\to B$
Câu 3:
Ta có:
$\displaystyle\int f(x)dx=\displaystyle\int \dfrac{\sin x}{1+3\cos x}dx$
$\to \displaystyle\int f(x)dx=\displaystyle\int \dfrac{-(-\sin x)}{1+3\cos x}dx$
$\to \displaystyle\int f(x)dx=\displaystyle\int \dfrac{-1}{1+3\cos x}d(\cos x)$
$\to \displaystyle\int f(x)dx=-\dfrac13\displaystyle\int \dfrac{3}{1+3\cos x}d(\cos x)$
$\to \displaystyle\int f(x)dx=-\dfrac13\displaystyle\int \dfrac{1}{1+3\cos x}d(3\cos x)$
$\to \displaystyle\int f(x)dx=-\dfrac13\ln|1+3\cos x|+C$
$\to F(x)=-\dfrac13\ln|1+3\cos x|+C$
Mà $F(\dfrac{\pi}{2})=2$
$\to -\dfrac13\ln|1+3\cdot \cos \dfrac{\pi}{2}|+C=2$
$\to C=2$
$\to F(x)=-\dfrac13\ln|1+3\cos x|+2$
$\to F(0)=-\dfrac13\ln|1+3\cos 0|+2=-\dfrac23\ln2+2$
$\to B$
Câu 4:
Ta có:
$\displaystyle\int \cos x\sin^2xdx$
$=\displaystyle\int \sin^2xd(\sin x)$
$=\dfrac{\sin^3x}{3}+C$
$\to C$
