Evaluate the integral. int(21x^(4)-3x^(2)-6)/(x^(2))dx Answer:

Simplify integrand: We are given the integral: int((21x^4 - 3x^2 - 6) / x^2)dx First, we simplify the integrand by dividing each term by x^2. (21x^4) / x^2 = 21x^(4-2) = 21x^2 (-3x^2) / x^2 = -3x^(2-2) = -3 (-6) / x^2 = -6x^(-2) So the integral becomes: int(21x^2 - 3 - 6x^(-2))dxIntegrate terms separately: Now we integrate each term separately. The integral of 21x^2 with respect to x is (21/3)x^(2+1) = 7x^3. The integral of -3 with respect to x is -3x. The integral of -6x^(-2) with respect to x is -6(-1)x^(-2+1) = 6x^(-1) = 6/x. So the integral becomes: 7x^3 - 3x + 6/x + C, where C is the constant of integration.

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Find indefinite integrals using the substitution

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Q. Evaluate the integral.

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\int \frac{21 x^{4}-3 x^{2}-6}{x^{2}} \mathrm{~d} x

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Answer:

Simplify integrand: We are given the integral: $\newline$ $\int(\frac{21x^4 - 3x^2 - 6}{x^2})dx$ $\newline$ First, we simplify the integrand by dividing each term by $x^2$ . $\newline$ $(\frac{21x^4}{x^2}) = 21x^{(4-2)} = 21x^2$ $\newline$ $(\frac{-3x^2}{x^2}) = -3x^{(2-2)} = -3$ $\newline$ $(\frac{-6}{x^2}) = -6x^{(-2)}$ $\newline$ So the integral becomes: $\newline$ $\int(21x^2 - 3 - 6x^{-2})dx$
Integrate terms separately: Now we integrate each term separately. $\newline$ The integral of $21x^2$ with respect to $x$ is $(\frac{21}{3})x^{(2+1)} = 7x^3$ . $\newline$ The integral of $-3$ with respect to $x$ is $-3x$ . $\newline$ The integral of $-6x^{-2}$ with respect to $x$ is $-6(-1)x^{-2+1} = 6x^{-1} = \frac{6}{x}$ . $\newline$ So the integral becomes: $\newline$ $7x^3 - 3x + \frac{6}{x} + C$ , where $x$ $0$ is the constant of integration.