allery$\int \dfrac{dx}{x^2+2x+2}$ equals
A$x\tan^{-1}(x+1)+C$
B$\tan^{-1}(x+1)+C$
C$(x+1)\tan^{-1}x+C$
D$\tan^{-1}x+C$
Answer & Solution
Correct answer: B. $\tan^{-1}(x+1)+C$
1. Complete the square: $x^2+2x+2=(x+1)^2+1$.
2. Put $t=x+1$, $dt=dx$.
3. Integral $=\int \dfrac{dt}{t^2+1}=\tan^{-1}t$.
4. Back-substitute: $\tan^{-1}(x+1)+C$. Differentiating confirms B.
_Source: syllabus Mathematics Ch 7 "Integrals", p.314_
Keep practising Probability & Statistics
Browse all Probability & Statistics questions for AP Intermediate 1st Year, or explore more Mathematics topics.
Related questions
$\int x\cos x\,dx$ equalsThe integration-by-parts formula $\int u\,\dfrac{dv}{dx}\,dx$ equals$\int \dfrac{dx}{x(x^2+1)}$ equals$\int \dfrac{x\,dx}{(x-1)(x-2)}$ equalsIn decomposing $\dfrac{1}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}$, the constants are$\int \dfrac{dx}{(x+1)(x+2)}$ equals$\int \dfrac{dx}{x^2-6x+13}$ equals$\int \dfrac{dx}{ qrt{2x-x^2}}$ equals