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Pascal Triangle

If ∫x4+3x+2 d...

Question

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

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Solution

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

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