Question

tìm nguyên hàm F(x) của hàm số f(x) thỏa mãn điều kiện tương ứng

a) \(f\left(x\right)=\dfrac{3}{x}-5x\) và \(F\left(e\right)=-1\)

b) \(f\left(x\right)=\dfrac{1}{x}-2x\) và \(F\left(e\right)=1\)

Answer 1

a: \(f\left(x\right)=\dfrac{3}{x}-5x\)

=>\(F\left(x\right)=\int\left(\dfrac{3}{x}-5x\right)dx=3ln\left|x\right|-5\cdot\dfrac{1}{2}x^2+C\)

=>\(F\left(x\right)=3\cdot ln\left|x\right|-\dfrac{5}{2}x^2+C\)

F(e)=-1

=>\(3\cdot ln\left|e\right|-\dfrac{5}{2}e^2+C=-1\)

=>\(C=-1-3\cdot ln\left|e\right|+\dfrac{5}{2}e^2=-4+\dfrac{5}{2}e^2\)

=>\(F\left(x\right)=3ln\left|x\right|-\dfrac{5}{2}x^2-4+\dfrac{5}{2}e^2\)

b: \(f\left(x\right)=\dfrac{1}{x}-2x\)

=>\(F\left(x\right)=\int\left(\dfrac{1}{x}-2x\right)dx=ln\left|x\right|-2\cdot\dfrac{1}{2}x^2+C=ln\left|x\right|-x^2+C\)

F(e)=1

=>\(ln\left|e\right|-e^2+C=1\)

=>\(C=e^2\)

=>F(x)=ln|x|-x²+e²