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Question Number 27282 by hp killer last updated on 04/Jan/18

$$\int log(2+x^2)dx$$

Commented by abdo imad last updated on 04/Jan/18

$$ifyoumeanlnintegrateparparts{u}^{{}^{\prime }}=1andv=ln(2+x^2)$$$$\int ln(2+x^2)dx=xln(2+x^2)-\int x\frac{2x}{2+x^2}dx$$$$=xln(2+x^2)-2\int \frac{x^2}{2+x^2}dx$$$$=xln(2+x^2)-2\int \frac{2+x^2-2}{2+x^2}dx$$$$=xln(2+x^2)-2x+4\int \frac{dx}{2+x^2}andbythechangementx=\sqrt{2}t$$$$\int \frac{dx}{2+x^2}=\int \frac{\sqrt{2}dt}{2+2t^2}=\frac{\sqrt{2}}{2}artan\left(\frac{x}{\sqrt{2}}\right)so$$$$\int ln(2+x^2)dx=xln(2+x^2)-2x+2\sqrt{2}arctan\left(\frac{x}{\sqrt{2}}\right).$$

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