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Differentiate w.r.t. x
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\int \left(8\left(x^{2}\right)^{3}+12\left(x^{2}\right)^{2}+6x^{2}+1\right)x\mathrm{d}x
Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(2x^{2}+1\right)^{3}.
\int \left(8x^{6}+12\left(x^{2}\right)^{2}+6x^{2}+1\right)x\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.
\int \left(8x^{6}+12x^{4}+6x^{2}+1\right)x\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.
\int 8x^{7}+12x^{5}+6x^{3}+x\mathrm{d}x
Use the distributive property to multiply 8x^{6}+12x^{4}+6x^{2}+1 by x.
\int 8x^{7}\mathrm{d}x+\int 12x^{5}\mathrm{d}x+\int 6x^{3}\mathrm{d}x+\int x\mathrm{d}x
Integrate the sum term by term.
8\int x^{7}\mathrm{d}x+12\int x^{5}\mathrm{d}x+6\int x^{3}\mathrm{d}x+\int x\mathrm{d}x
Factor out the constant in each of the terms.
x^{8}+12\int x^{5}\mathrm{d}x+6\int x^{3}\mathrm{d}x+\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 8 times \frac{x^{8}}{8}.
x^{8}+2x^{6}+6\int x^{3}\mathrm{d}x+\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 12 times \frac{x^{6}}{6}.
x^{8}+2x^{6}+\frac{3x^{4}}{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}.
x^{8}+2x^{6}+\frac{3x^{4}}{2}+\frac{x^{2}}{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}.
\frac{x^{2}}{2}+\frac{3x^{4}}{2}+2x^{6}+x^{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.