Integrate w.r.t to $x$
$3 x + 1$

\frac{3 x^{2}}{2} + x + c

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x^{3} + 3 x + c

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3 x + c

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None of these

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Solution

The correct option is B $\frac{3 x^{2}}{2} + x + c$
$\int (3 x + 1) d x$

$= 3 \int x d x + \int d x$

$= \frac{{3 x}^{2}}{2} + x + c$

Suggest Corrections

Similar questions

\int \frac{(x^{3} + 3 x^{2} + 3 x + 1)}{(x + 1)^{5}} = d x

\int \frac{{(x - 1)}^{2}}{x^{3} - 3 x^{2} + 3 x + 8} d x

\frac{d y}{d x} - \frac{y + 3 x}{{log}_{e} (y + 3 x)} + 3 = 0

c

x^{3} - 3 x^{2} + 3 x - 7 = (x + 1) (a x^{2} + b x + c)

a + b + c

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