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Integral Calculus> indefinite integration...

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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

Manoj Kumar Jangra , 16 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 11 Years ago

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

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