Question-137461

Question Number 137461 by mathlove last updated on 03/Apr/21

Answered by Ñï= last updated on 03/Apr/21

\int \frac{1 + x^{2}}{2 x + x^{3}} d x = \frac{1}{3} \int \frac{2 + 3 x^{2} + 1}{2 x + x^{3}} d x = \frac{1}{3} l n ∣ 2 x + x^{3} ∣ + \frac{1}{3} \int \frac{d x}{x (2 + x^{2})}

= \frac{1}{3} l n ∣ 2 x + x^{3} ∣ + \frac{1}{6} \int (\frac{1}{x} - \frac{x}{2 + x^{2}}) = \frac{1}{3} l n ∣ 2 x + x^{3} ∣ + \frac{1}{6} l n ∣ x ∣ - \frac{1}{12} l n ∣ 2 + x^{2} ∣ + C

Answered by MJS_new last updated on 03/Apr/21

\int \frac{x^{2} + 1}{x (x^{2} + 2)} d x = \frac{1}{2} \int \frac{d x}{x} + \frac{1}{2} \int \frac{x}{x^{2} + 2} =

= \frac{1}{2} \ln ∣ x ∣ + \frac{1}{4} \ln (x^{2} + 2) + C

Answered by liberty last updated on 04/Apr/21

\int \frac{x^{2} + 1}{x (2 + x^{2})} d x = \int \frac{x}{2 + x^{2}} d x + \int \frac{1}{x (2 + x^{2})} d x

= \frac{1}{2} \int \frac{d (2 + x^{2})}{2 + x^{2}} + \int \frac{\sqrt{2} \sec^{2} t}{\sqrt{2} \tan t (2 \sec^{2} t)} d t

[x = \sqrt{2} \tan t]

I = \frac{1}{2} \ln (2 + x^{2}) + \frac{1}{2} \int \frac{\cos t}{\sin t} d t

I = \frac{1}{2} \ln (2 + x^{2}) + \frac{1}{2} \ln (\sin t) + c

I = \frac{1}{2} \ln (2 + x^{2}) + \frac{1}{2} \ln (\frac{∣ x ∣}{\sqrt{2 + x^{2}}}) + c

Commented by MJS_new last updated on 04/Apr/21

\frac{1}{2} \ln a + \frac{1}{2} \ln \frac{b}{\sqrt{a}} = \frac{1}{2} \ln a + \frac{1}{2} \ln b - \frac{1}{2} \ln \sqrt{a} =

= \frac{1}{2} \ln a + \frac{1}{2} \ln b - \frac{1}{4} \ln a = \frac{1}{4} \ln a + \frac{1}{2} \ln b

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