Question
Solve: $\int\frac{\text{x}^2+1}{\text{x}^2+1}\text{dx}=$
- 1 + C
- x2 + C
- x + C
- 0
Solution:
Now, $\int\frac{\text{x}^2+1}{\text{x}^2+1}\text{dx}$
$=\int\text{dx}$
= x + C [Where C is integrating constant]
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$(a-c) x^2+(b-a) x+(c-b)=0$ where $a, b, c$ are distinct real numbers such that the matrix
$\left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\1 & 1 & 1 \\a & b & c\end{array}\right]$
is singular. Then the value of
$\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$