logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

integrate 1/1+x3

Can you answer this question?
 
 

1 Answer

0 votes
<div class="clay6-step-odd"><div class="clay6-basic" id="pr10">Ans:  $-\frac{1}{6} log(x^2 - x + 1 ) + \frac{1}{3} log(x+1) + \frac{1}{\sqrt{3}} tan^{-1} ( \frac{2x-1}{\sqrt{3}} )$</div></div><div class="clay6-step-even"><div class="clay6-basic" id="pr20">$1+x^3=(1+x)(1-x+x^2)$</div><div class="clay6-basic" id="pr21">$\large\frac{1}{1+x^3}=\frac{1}{(1+x)(1-x+x^2)}$</div><div class="clay6-basic" id="pr22">Let  $\large\frac{1}{(1+x)(1-x+x^2)}=\frac{A}{1+x}$$+\large\frac{Bx+C}{1-x+x^2}$</div><div class="clay6-basic" id="pr23">$\Rightarrow\:1=A(1-x+x^2)+(Bx+C)(1+x)$</div><div class="clay6-basic" id="pr24">Put   $x=-1$</div><div class="clay6-basic" id="pr25">$\Rightarrow\:1=3A$  or  $A=\large\frac{1}{3}$</div><div class="clay6-basic" id="pr26">Put $x=0$</div><div class="clay6-basic" id="pr27">$\Rightarrow\:1=A+C$ </div><div class="clay6-basic" id="pr28">Substituting the value of $A$  we get $C=1-\large\frac{1}{3}=\frac{2}{3}$</div><div class="clay6-basic" id="pr29">Put  $x=1$</div><div class="clay6-basic" id="pr210">$\Rightarrow\:1=A+2B+2C$</div><div class="clay6-basic" id="pr211">Substituting the values of $A\:and\:C$  we get  </div><div class="clay6-basic" id="pr212">$2B=1-\large\frac{1}{3}-\frac{4}{3}=0$   or   $B=-\large\frac{1}{3}$</div></div><div class="clay6-step-odd"><div class="clay6-basic" id="pr30">Substituting the values of $A,B,C$  we have</div><div class="clay6-basic" id="pr31">$\large\frac{1}{1+x^3}=\frac{1}{(1+x)(1-x+x^2)}=\frac{1}{3}.\frac{1}{1+x}$$+\large\frac{\large\frac{-1}{3}.x+\frac{2}{3}}{1-x+x^2}$</div><div class="clay6-basic" id="pr32">$\therefore\:\int\:\large\frac{1}{1+x^3}$$dx=\large\frac{1}{3}$$\int\:\large\frac{dx}{1+x}$$+\large\frac{1}{3}.$$\int\:\large\frac{-x+2}{1-x+x^2}$$dx$</div><div class="clay6-advanced" id="pr33">$\large\frac{dx}{1+x}$$=log|x+1|$</div><div class="clay6-basic" id="pr34">$\large\frac{1}{3}.$$\int\:\large\frac{-x+2}{1-x+x^2}$$dx=-\large\frac{1}{6}$$\int\large\frac{2x-1-3}{1-x+x^2}$$dx$</div><div class="clay6-basic" id="pr35">$=-\large\frac{1}{6}$$\bigg[\int\large\frac{2x-1}{1-x+x^2}$$dx-3\int\:\large\frac{dx}{1-x+x^2}\bigg]$</div><div class="clay6-advanced" id="pr36">$\int\large\frac{2x-1}{1-x+x^2}$$dx=log|1-x+x^2|$</div><div class="clay6-basic" id="pr37">$\int\large\frac{dx}{1-x+x^2}$$=\int\:\large\frac{dx}{(x-1/2)^2+\large\frac{3}{4}}$</div><div class="clay6-advanced" id="pr38">$\int\large\frac{dx}{x^2+a^2}$$=\large\frac{1}{a}$$tan^{-1}(\large\frac{x}{a})$</div><div class="clay6-basic" id="pr39">$\therefore\:\int\:\large\frac{dx}{(x-1/2)^2+\large\frac{3}{4}}$$=\large\frac{2}{\sqrt 3}$$tan^{-1}\large\frac{2(x-\large\frac{1}{2})}{\sqrt 3}$</div><div class="clay6-basic" id="pr310">$=\large\frac{2}{\sqrt 3}.$$tan^{-1}\large\frac{2x-1}{\sqrt 3}$</div><div class="clay6-basic" id="pr311">$\Rightarrow\:-\large\frac{1}{6}$$\bigg[\int\large\frac{2x-1}{1-x+x^2}$$dx-3\int\:\large\frac{dx}{1-x+x^2}\bigg]$</div><div class="clay6-basic" id="pr312">$\therefore\:\int\:\large\frac{1}{1+x^3}$$dx=\large\frac{1}{3}$$log|x+1|-\large\frac{1}{6}$$log|1-x+x^2|+\large\frac{1}{\sqrt 3}$$tan^{-1}\large\frac{2x-1}{\sqrt 3}$</div></div>
answered Feb 15, 2014 by balaji
edited Feb 16, 2014 by rvidyagovindarajan_1
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...