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

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

$$\int \frac{1+x^2}{2x+x^3}dx=\frac{1}{3}\int \frac{2+3x^2+1}{2x+x^3}dx=\frac{1}{3}ln\mid 2x+x^3\mid +\frac{1}{3}\int \frac{dx}{x(2+x^2)}$$$$=\frac{1}{3}ln\mid 2x+x^3\mid +\frac{1}{6}\int (\frac{1}{x}-\frac{x}{2+x^2})=\frac{1}{3}ln\mid 2x+x^3\mid +\frac{1}{6}ln\mid x\mid -\frac{1}{12}ln\mid 2+x^2\mid +C$$

Answered by MJS_new last updated on 03/Apr/21

$$\int \frac{x^2+1}{x(x^2+2)}dx=\frac{1}{2}\int \frac{dx}{x}+\frac{1}{2}\int \frac{x}{x^2+2}=$$$$=\frac{1}{2}\ln \mid x\mid +\frac{1}{4}\ln (x^2+2)+C$$

Answered by liberty last updated on 04/Apr/21

$$\int \frac{x^2+1}{x(2+x^2)}dx=\int \frac{x}{2+x^2}dx+\int \frac{1}{x(2+x^2)}dx$$$$=\frac{1}{2}\int \frac{d(2+x^2)}{2+x^2}+\int \frac{\sqrt{2}\sec {}^{2}t}{\sqrt{2}\tan t\left(2\sec {}^{2}t\right)}dt$$$$[x=\sqrt{2}\tan t]$$$$I=\frac{1}{2}\ln (2+x^2)+\frac{1}{2}\int \frac{\cos t}{\sin t}dt$$$$I=\frac{1}{2}\ln (2+x^2)+\frac{1}{2}\ln \left(\sin t\right)+c$$$$I=\frac{1}{2}\ln (2+x^2)+\frac{1}{2}\ln \left(\frac{\mid x\mid }{\sqrt{2+x^2}}\right)+c$$

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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