int frac{1}{x^2} (2x + 1)^3 dx = | vedclass

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(A) To solve the integral $\int \frac{1}{x^2} (2x + 1)^3 dx$,we first expand the binomial $(2x + 1)^3$.Using the formula $(a+b)^3 = a^3 + 3a^2b + 3ab^

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

Vedclass · Answer

(A) To solve the integral $\int \frac{1}{x^2} (2x + 1)^3 dx$,we first expand the binomial $(2x + 1)^3$.Using the formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we get $(2x + 1)^3 = (2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + (1)^3 = 8x^3 + 12x^2 + 6x + 1$.Now,substitute this into the integral: $\int \frac{8x^3 + 12x^2 + 6x + 1}{x^2} dx$.Divide each term by $x^2$: $\int (8x + 12 + \frac{6}{x} + \frac{1}{x^2}) dx$.Integrate term by term: $\int 8x dx + \int 12 dx + \int \frac{6}{x} dx + \int x^{-2} dx$.This results in $8 \cdot \frac{x^2}{2} + 12x + 6\log|x| + \frac{x^{-1}}{-1} + c$.Simplifying,we get $4x^2 + 12x + 6\log|x| - \frac{1}{x} + c$.

$\int \frac{1}{x^2} (2x + 1)^3 dx = $

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