$\int \dfrac{1}{x}\,dx$ equals
A$-\dfrac{1}{x^2}+C$
B$x\log|x|+C$
C$\dfrac{x^2}{2}+C$
D$\log|x|+C$
Answer & Solution
Correct answer: D. $\log|x|+C$
1. The power rule fails for $n=-1$.
2. Recall $\dfrac{d}{dx}(\log|x|)=\dfrac{1}{x}$ for $x\neq 0$.
3. Hence $\int \dfrac{1}{x}\,dx=\log|x|+C$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.290_
Related questions
$\displaystyle\int_2^3 \dfrac{x\,dx}{x^2+1}$ equals$\displaystyle\int_0^{2a} f(x)\,dx$ equals $2\displaystyle\int_0^{a} f(x)\,dx$ precisely wIf $f(a+b-x)=f(x)$, then $\displaystyle\int_a^b x\,f(x)\,dx$ is equal to$\displaystyle\int \frac{dx}{e^{x}+e^{-x}}$ is equal toA rational function $\dfrac{P(x)}{Q(x)}$ is called proper when$\displaystyle\int_0^{\pi/4} \tan x\,dx$ equals$\displaystyle\int_0^{1} x e^{x^2}\,dx$ equalsThe value of $\displaystyle\int_0^{\pi/2} \log\!\left(\dfrac{4+3 in x}{4+3\cos x}\right)dx