$\int \dfrac{2x}{1+x^2}\,dx$ equals
A$\tan^{-1}x+C$
B$2\log(1+x^2)+C$
C$\dfrac{1}{1+x^2}+C$
D$\log(1+x^2)+C$
Answer & Solution
Correct answer: D. $\log(1+x^2)+C$
1. Put $t=1+x^2$, so $dt=2x\,dx$ (the numerator).
2. Integral $=\int \dfrac{dt}{t}=\log|t|$.
3. Since $1+x^2>0$: $\log(1+x^2)+C$.
4. Option A would need numerator $1$, not $2x$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.302_
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