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Manoj Kumar Jangra Grade: 12
        1. integration of(1+cos x/n)*dx

* = means square root
2. integration of x^4/1+x^2dx
^2 means raise to the power 2
^4 means raise to the power 4
3. integration of dx/(1+x^4)^1/4
^1/4 means raise to the power 1/4
4. integration of (t^3 -1)*/t dt
* means square root
7 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
										
Ans:
I_{1} = \int (1+cos\frac{x}{n})dx
I _{1}= x + nsin(\frac{x}{n}) + constant
I _{2}= \int \frac{x^{4}}{x^{2}+1} dx
I _{2}= \int \frac{x^{4}-1+1}{x^{2}+1} dx
I _{2}= \int( x^{2}+\frac{1}{x^{2}+1}-1) dx
I _{2}= \frac{x^{3}}{3} + tan^{-1}x-x+constant
I _{3}= \int \frac{1}{x^{4}+1}dx
Simply use the partial fraction rule here, we have
I _{3}= \int (\frac{\sqrt{2}x-2}{4(-x^{2}+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^{2}+\sqrt{2}x+1)})dx
I _{3}= \frac{log(\frac{x^{2}+\sqrt{2}x+1}{x^{2}-\sqrt{2}x+1})+2tan^{-1}(\sqrt{2}x+1)-2tan^{-1}(1-\sqrt{2}x)}{4\sqrt{2}}+constant
I _{4}= \int \frac{t^{3}-1}{t}dt
I _{4}= \int (t^{2}-\frac{1}{t})dt
I _{4}= \frac{t^{3}}{3} - ln(t) + constant
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
3 years ago
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