Question

$∫\dfrac{x+5}{x^{2}+1}dx=$

• A. $3ln|x-1|-2ln|x+1|+C$
• B. $2ln|x-1|-3ln|x+1|+C$
• C. $ln|x-2|-ln|x+1|+C$
• D. $ln|x+2|+ln|x+1|+C$
• E. $2ln|x-1|+3ln|x-1|+C$

Answer 1

Answer: (A)

$3ln|x-1|-2ln|x+1|+C$

Answer 2

$∫\dfrac{x+5}{x^{2}+1}dx=$

So to solve this question, we can again use partial fraction decomposition.

_Step 1: _

Factorize the denominator. $(x^2 - 1)$ can be factored as$(x - 1)(x + 1).$

Step 2:

Partial fraction decomposition. The expression $\dfrac{x+5}{x^2 - 1}$ can be rewritten as the sum of two fractions with unknown constants $A$ and $B$:

$\dfrac{x+5}{x^2 - 1} = \dfrac{A}{x +1}+\dfrac{B}{x-1}$

Step 3 :

Now to find the values of A and B, we need to find a common denominator, which is $(x- 1)(x + 1)$, and then equate the numerators:

$x + 5 = A(x+ 1) + B(x- 1)$

Now, solve for A and B by comparing coefficients: $A + B = 1$

(by comparing the coefficients of $x$) $A - B = 5$ ⇢(by comparing the constant terms)

Adding the two equations: $2A = 6$

                                        ⇒$A=3$

Substituting the value of $A$ one of the equations to find we get

$B = -2$

_Step 4: _

Now we can re-write the parent expression as,

$∫\dfrac{x + 5}{x^2 - 1} dx = ∫(\dfrac{A}{x - 1}+\dfrac{B}{x+ 1}) dx$

$=∫(\dfrac{3}{x-1}) - (\dfrac{2}{x+1})dx$

$=∫(\dfrac{3}{x-1}) - (\dfrac{2}{x+1})dx$

$=∫(\dfrac{3}{x-1})dx - ∫(\dfrac{2}{x+1})dx$

$= 3ln|x-1|-2ln|x+1|+C$ (Ans.)