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

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Question

(A) To evaluate the integral $\int \frac{2x}{(2x + 1)^2} dx$,we rewrite the numerator as $(2x + 1 - 1)$.$\int \frac{2x + 1 - 1}{(2x + 1)^2} dx = \int

Vedclass · Accepted Answer

$\frac{1}{2}\log(2x + 1) + \frac{1}{2(2x + 1)} + c$

Vedclass · Answer

(A) To evaluate the integral $\int \frac{2x}{(2x + 1)^2} dx$,we rewrite the numerator as $(2x + 1 - 1)$.$\int \frac{2x + 1 - 1}{(2x + 1)^2} dx = \int \left( \frac{2x + 1}{(2x + 1)^2} - \frac{1}{(2x + 1)^2} \right) dx$$= \int \frac{1}{2x + 1} dx - \int (2x + 1)^{-2} dx$Using the standard integration formulas $\int \frac{1}{ax+b} dx = \frac{1}{a}\log|ax+b| + c$ and $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$:$= \frac{1}{2}\log|2x + 1| - \frac{(2x + 1)^{-1}}{2(-1)} + c$$= \frac{1}{2}\log(2x + 1) + \frac{1}{2(2x + 1)} + c$.

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

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