Mathematics · 2019
JEE Main · 10 April 2019, Shift 1 · Q74
If ∫ (d x)/((x^2-2 x+10)^2); = A ( tan^-1((x-1)/3)+(f(x))/(x^2-2 x+10))+ C where C is a constant of integration, then:
$$\begin{aligned}
& \text { If } \int \frac{\mathrm{d} x}{\left(x^2-2 x+10\right)^2} \\
& =\mathrm{A}\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{f(x)}{x^2-2 x+10}\right)+\mathrm{C}
\end{aligned}
$$
where $\displaystyle C$ is a constant of integration, then :
Official answer
(3)
$\displaystyle \mathrm{A}=\frac{1}{54} \text { and } f(x)=3(x-1)$
More from Indefinite Integration
- If ∫ 1/x √((1-x)/(1+x)) d x=g(x)+c, g(1)=0, then g(1/2) is equal to:2022
- If ∫ (cos x d x)/(sin^3 x(1+ sin^6 x)^2 / 3)=f(x)(1+ sin^6 x)^1 / λ+c where c is a constant of integration, then λ f((π)/3) is equal to:2020
- Let f(x)=∫ (d x)/((3+4 x^2) √(4-3 x^2)),|x|<2/(√3). If f(0)=0 and f(1)=1/(α β) tan^-1((α)/(β)), α, β>0, then α^2+β^2 is equal to ____.2023
- If ∫ √(sec 2 x-1) d x=α log_e| cos 2 x+β+√(cos 2 x(1+ cos 1/(β) x))|+ constant, then β-α is equal to ____.2023
- The integral ∫(x/(x sin x+ cos x))^2 d x is equal to (where C is a constant of integration):2020
- Let I(x)=∫ (x^2(x sec^2 x+ tan x))/((x tan x+1)^2) d x. If I(0)=0, then I((π)/4) is equal to2023
- The integral ∫ ((x^8-x^2) d x)/((x^12+3 x^6+1) tan^-1(x^3+1/x^3)) is equal to:2024
- If ∫ sin^-1(√(x/(1+x))) d x=A(x) tan^-1(√x)+ B(x)+C, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be:2020
JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.