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\int x^{3}+4x+9\mathrm{d}x
Evaluate the indefinite integral first.
\int x^{3}\mathrm{d}x+\int 4x\mathrm{d}x+\int 9\mathrm{d}x
Integrate the sum term by term.
\int x^{3}\mathrm{d}x+4\int x\mathrm{d}x+\int 9\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{4}}{4}+4\int x\mathrm{d}x+\int 9\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.
\frac{x^{4}}{4}+2x^{2}+\int 9\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. 4 কে \frac{x^{2}}{2} বার গুণ করুন।
\frac{x^{4}}{4}+2x^{2}+9x
Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{10^{4}}{4}+2\times 10^{2}+9\times 10-\left(\frac{2^{4}}{4}+2\times 2^{2}+9\times 2\right)
The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.
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