int frac{1}{x^2}(2 x+1)^3 d x is equal to | vedclass

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(A) We have,$\int \frac{1}{x^2}(2 x+1)^3 d x$ Expanding the numerator using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$: $(2x+1)^3 = (2x)^3 + 3(2x)^2(1) + 3

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$4 x^2+12 x+6 \log |x|-\frac{1}{x}+C$

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(A) We have,$\int \frac{1}{x^2}(2 x+1)^3 d x$ Expanding the numerator using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$: $(2x+1)^3 = (2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + (1)^3 = 8x^3 + 12x^2 + 6x + 1$ Now,the integral becomes: $\int \frac{8x^3 + 12x^2 + 6x + 1}{x^2} d x$ $= \int (8x + 12 + \frac{6}{x} + \frac{1}{x^2}) d x$ Integrating term by term: $= 8 \int x d x + 12 \int d x + 6 \int \frac{1}{x} d x + \int x^{-2} d x$ $= 8(\frac{x^2}{2}) + 12x + 6 \log |x| + (\frac{x^{-1}}{-1}) + C$ $= 4x^2 + 12x + 6 \log |x| - \frac{1}{x} + C$

$\int \frac{1}{x^2}(2 x+1)^3 d x$ is equal to

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