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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=\sin x+x\ln x$ là
A. $F\left( x \right)=-\cos x+\ln x+C$
B. $F\left( x \right)=-\cos x+\dfrac{{{x}^{2}}}{2}\ln x-\dfrac{{{x}^{2}}}{4}+C$
C. $F\left( x \right)=\cos x+\dfrac{{{x}^{2}}}{2}\ln x-\dfrac{{{x}^{2}}}{4}+C$
D. $F\left( x \right)=-\cos x+C$
A. $F\left( x \right)=-\cos x+\ln x+C$
B. $F\left( x \right)=-\cos x+\dfrac{{{x}^{2}}}{2}\ln x-\dfrac{{{x}^{2}}}{4}+C$
C. $F\left( x \right)=\cos x+\dfrac{{{x}^{2}}}{2}\ln x-\dfrac{{{x}^{2}}}{4}+C$
D. $F\left( x \right)=-\cos x+C$
$\int{\left( \sin x+x\ln x \right)dx}=-\cos x+\int{x\ln x}=-\cos x+\dfrac{1}{2}\int{\ln xd{{x}^{2}}}$
$=-\cos x+\dfrac{{{x}^{2}}}{2}.\ln x-\dfrac{1}{2}\int{xdx}=-\cos x+\dfrac{{{x}^{2}}}{2}.\ln x-\dfrac{{{x}^{2}}}{4}+C$.
$=-\cos x+\dfrac{{{x}^{2}}}{2}.\ln x-\dfrac{1}{2}\int{xdx}=-\cos x+\dfrac{{{x}^{2}}}{2}.\ln x-\dfrac{{{x}^{2}}}{4}+C$.
Đáp án B.