Question

If $\underset{}{\int }(x^4+3x+2)dx=\frac{x^5}{a}+\frac{3x^2}{b}+cx+K$, then the value of $a+b+c$ is (where $K$ is constant of integration and $a,b$ and $c$ are fixed constants)

• A. 9
• B. 9.00
• C. 9.0
• D. 09

Answer 1

Answer: (A), (B), (C), (D)

$\int (x^4+3x+2)dx=\int x^{4}dx+\underset{}{\int }3xdx+\underset{}{\int }2dx$
$=\frac{x^5}{5}+\frac{3x^2}{2}+2x+K$
$\Rightarrow \frac{x^5}{a}+\frac{3x^2}{b}+cx+K=\frac{x^5}{5}+\frac{3x^2}{2}+2x+K$
By comparing, we get
$\Rightarrow a=5,b=2$ and $c=2$
$\therefore a+b+c=9$