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Differentiate w.r.t. x
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\int 2x\left(\left(x^{2}\right)^{3}+3\left(x^{2}\right)^{2}+3x^{2}+1\right)\mathrm{d}x
Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x^{2}+1\right)^{3}.
\int 2x\left(x^{6}+3\left(x^{2}\right)^{2}+3x^{2}+1\right)\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.
\int 2x\left(x^{6}+3x^{4}+3x^{2}+1\right)\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.
\int 2x^{7}+6x^{5}+6x^{3}+2x\mathrm{d}x
Use the distributive property to multiply 2x by x^{6}+3x^{4}+3x^{2}+1.
\int 2x^{7}\mathrm{d}x+\int 6x^{5}\mathrm{d}x+\int 6x^{3}\mathrm{d}x+\int 2x\mathrm{d}x
Integrate the sum term by term.
2\int x^{7}\mathrm{d}x+6\int x^{5}\mathrm{d}x+6\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{8}}{4}+6\int x^{5}\mathrm{d}x+6\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{7}\mathrm{d}x with \frac{x^{8}}{8}. Multiply 2 times \frac{x^{8}}{8}.
\frac{x^{8}}{4}+x^{6}+6\int x^{3}\mathrm{d}x+2\int x\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}. Multiply 6 times \frac{x^{6}}{6}.
\frac{x^{8}}{4}+x^{6}+\frac{3x^{4}}{2}+2\int x\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}. Multiply 6 times \frac{x^{4}}{4}.
\frac{x^{8}}{4}+x^{6}+\frac{3x^{4}}{2}+x^{2}
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}. Multiply 2 times \frac{x^{2}}{2}.
x^{2}+\frac{3x^{4}}{2}+x^{6}+\frac{x^{8}}{4}+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.