Question

$log(x ) = \frac{x - 1}{x + 1} + \frac{1}{2} \frac{ {x {}^{2} } - 1{} }{(x + 1) {}^{2} } + \frac{1}{3} \frac{ {x}^{3 - 1} }{(x + 1) {}^{3} } + ...$
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Answer 1

Answer:

To integrate the rationalrational function x(x−1)(x−2)(x−3)x(x−1)(x−2)(x−3)

Let integrand x(x−1)(x−2)(x−3)=Ax−1+Bx−2+Cx−3…(i)x(x−1)(x−2)(x−3)=Ax−1+Bx−2+Cx−3…(i) PartialfractionsPartialfractions

Multiplying by L. C.M. =(x−1)(x−2)(x−3) , x=Λ(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)=A(x2−5x+6)+B(x2−4x+3)+C(x2−3x+2)=Ax2−6Ax+6A+Bx2−4Bx+3B+Cx2−3Cx+2C

Multiplying by L. C.M. =(x−1)(x−2)(x−3) , x=Λ(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)=A(x2−5x+6)+B(x2−4x+3)+C(x2−3x+2)=Ax2−6Ax+6A+Bx2−4Bx+3B+Cx2−3Cx+2C

Comparing coefficients of x2,xx2,x and constant terms on both aides, we have x2x2 A+B+C=0A+B+C=0 xx ? −5A−4B−3C=1−5A−4B−3C=1 or 5A+4B+3C=−15A+4B+3C=−1

Constants: 6A+3B+2C=06A+3B+2C=0 Let us solve Eqns. (ii),(iii)(ii),(iii) and (iv)(iv) for A,B,C.A,B,C.

Let us first form two Eqns. in two unknowns say A and B.

Eqn. (ii)−3×(ii)−3× Eqn. iiii gives

5A+4B+3C−3A−3B−3C=−15A+4B+3C−3A−3B−3C=−1 or 2A+B=−12A+B=−1 …(v)…(v)

Eqn. (iv)−2×(iv)−2× Eqn. iiii gives

4A+B=04A+B=0

Eqn. (vi)−(vi)− Eqn. 00 gives ToeliminateBToeliminateB

2A=1∴A=122A=1∴A=12

Putting A=12A=12 in (v),1+B=−1⇒B=−2(v),1+B=−1⇒B=−2

Putting A=12A=12 and B=−2B=−2 in (ii)(ii)

12−2+C=0⇒C=−12+2=−1+42=3212−2+C=0⇒C=−12+2=−1+42=32

Putting these values of A, B, C in ii, we have

x(x−1)(x−2)(x−3)=12x−1−2x−2+32x−3

x(x−1)(x−2)(x−3)=12x−1−2x−2+32x−3

∴∫x(x−1)(x−2)(x−3)dx∴∫x(x−1)(x−2)(x−3)dx =12∫1x−1dx−2∫1x−2dx+32∫1x−3dx=12∫1x−1dx−2∫1x−2dx+32∫1x−3dx

=12log|x−1|−2log|x−2|+32log|x−3|+c

Answer 2

Step-by-step explanation:

$\mathfrak{\large{\green{\underline{\underline{solution}}}}} \\ \\ \large\implies{1 + 2 = 3}$