Question

(2x+5)dx
The value of f(x)=integral (2x+5)dx/
(x+1)(x+2)x+3)(x+4)+1
is (where C is constant of integration)
​

Answer 1

Answer:

Please see the attachment

Answer 2

Concept:

We need to have the knowledge of the below integration formula:

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

Where, c is constant of intergration.

This question also requires us to have knowledge on how to solve quadratic equation.

Given:

We are given the following intergration

$\int\limits {\frac{(2x+5}{(x+1)(x+2)(x+3)(x+4)+1} } \, dx$

To find:

We need to find the value of the above intergration.

Solution:

Lets solve the denominator.

$(x^{2}+5x+4)(x^{2} +5x+6)$

Let the integration be I

Therefore the intergration is

$I=\int\limits {\frac{(2x+5}{(x^{2}+5x+4)(x^{2} +5x+6)+1} } \, dx$

⇒$I=\int\limits {\frac{(2x+5}{(x^{2}+5x+4)((x^{2} +5x+4)+2)+1} } \, dx$

Let,

$x^{2}+5x+4=t\\ differentiating\ w.r.t\ x\\(2x+5)dx=dt\\dx=\frac{dt}{(2x+5)}$

Now substituting value of t and dx

$I=\int\ {\frac{2x+5}{t(t+2)+1}} \, \frac{dt}{(2x+5)}$

⇒$I=\int\ {\frac{1}{t(t+2)+1}} \,dt$

⇒$I=\int\ {\frac{1}{t^{2}+2t+1 } } \, dt=\int\ {(t+1)^{-2} } \, dt$     -----------(1)

Let,

t+1=v

differenciating w.r.t dv

$dt=dv$

Now substituting value of dt and v in equation (1)

$I=\int\ {v^{-2} } \, dv$

⇒$I=\frac{v^{-2+1} }{-2+1}+c=\frac{v^{-1} }{-1}+c$

⇒$I=-\frac{1}{v}+c$           -------------(2)

Now, we substitute value of v and t in equation (2)

Therefore,

$I=-\frac{1}{x^{2}+5x+4+1 }+c$

⇒$I=-\frac{1}{x^{2}+5x+5}+c$

Therefore, the value of the given integration is $-\frac{1}{x^{2}+5x+5}+c$.