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w.r.t. x পার্থক্য করুন
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শেয়ার করুন

\int 2x^{5}\mathrm{d}x+\int \frac{3}{x}\mathrm{d}x+\int \frac{1}{x^{9}}\mathrm{d}x
Integrate the sum term by term.
2\int x^{5}\mathrm{d}x+3\int \frac{1}{x}\mathrm{d}x+\int \frac{1}{x^{9}}\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{6}}{3}+3\int \frac{1}{x}\mathrm{d}x+\int \frac{1}{x^{9}}\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. 2 কে \frac{x^{6}}{6} বার গুণ করুন।
\frac{x^{6}}{3}+3\ln(|x|)+\int \frac{1}{x^{9}}\mathrm{d}x
Use \int \frac{1}{x}\mathrm{d}x=\ln(|x|) from the table of common integrals to obtain the result.
\frac{x^{6}}{3}+3\ln(|x|)-\frac{1}{8x^{8}}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{9}}\mathrm{d}x with -\frac{1}{8x^{8}}.
\frac{x^{6}}{3}+3\ln(|x|)-\frac{1}{8x^{8}}+С
If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.