Question

intregation on
xdx/(x + 1)(2x + 1)
​

Answer 1

Answer:

$(x + 1)(2x - 1)dx$

$= \frac{2}{3} {x}^{3} + \frac{1}{2} {x}^{2} - x + c$

Step-by-step explanation:

Find an antiderivative of a function is the same as finding the integral.

So we want to find

$(x + 1)(2x - 1)dx$

$2 {x}^{2} + x - 1dx$

To find this, we use the power rule

${x}^{n} dx = \frac{1}{n + 1} {x}^{n} + c$

$=> 2 {x}^{2} + x - 1dx$

$= \frac{2}{3} {x}^{3} + \frac{1}{2} {x}^{2} - x + c$

Hope This Helps ☺

Answer 2

Solution

We Have

$\sf \implies \: \int \dfrac{x}{(x + 1)(2x + 1)} dx$

Using practical Fraction Method

$\sf \implies \: \dfrac{x}{(x + 1)(2x + 1)} = \dfrac{A }{(x + 1)} + \dfrac{B}{(2x + 1)}$

$\sf \implies \: \dfrac{x}{(x + 1)(2x + 1)} = \dfrac{A(2x + 1) + B(x + 1)}{(x + 1)(2x + 1)}$

$\sf \implies \: \dfrac{x}{ \cancel{(x + 1)(2x + 1)}} = \dfrac{A(2x + 1) + B(x + 1)}{ \cancel{(x + 1)(2x + 1)}}$

$\sf \implies \: x = A(2x + 1) + B(x + 1)$

When x = - 1

$\sf \implies \: - 1 = A(2 \times - 1 + 1) +B ( - 1 + 1)$

$\sf \implies \: - 1 = A( - 1) + 0$

$\sf \implies \: A = 1$

When x = -1/2

$\sf \implies \: \dfrac{ - 1}{2} = A (2 \times \dfrac{ - 1}{2} + 1) + B \bigg( \dfrac{ - 1}{2} + 1 \bigg)$

$\sf \implies \: \dfrac{ - 1}{2} = 0 + B \dfrac{1}{2}$

$\sf \implies \: B = - 1$

Now put the value A = 1 and B = - 1

$\sf \implies \: \dfrac{x}{(x + 1)(2x + 1)} = \dfrac{A }{(x + 1)} + \dfrac{B}{(2x + 1)}$

$\sf \implies \: \dfrac{x}{(x + 1)(2x + 1)} = \dfrac{1 }{(x + 1)} + \dfrac{ - 1}{(2x + 1)}$

$\sf \implies \int \bigg( \dfrac{1 }{(x + 1)} + \dfrac{ - 1}{(2x + 1)} \bigg)dx$

$\sf \implies \: \int \dfrac{1}{(x + 1)} dx + \int \dfrac{ - 1}{(2x + 1)} dx$

$\sf \implies \: \int \dfrac{1}{(x + 1)} dx - \int \dfrac{ 1}{(2x + 1)} dx$

$\sf \implies \: ln(x + 1) - \dfrac{1}{2} ln(2x + 1) + c$

Answer

$\sf \implies \: ln(x + 1) - \dfrac{1}{2} ln(2x + 1) + c$