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## Homework Statement

[tex](x^{2}+1)(tan y)(y')=x[/tex]

## Homework Equations

## The Attempt at a Solution

[tex](x^{2}+1)(tan y)(\frac{dy}{dx})=x[/tex]

[tex](tan y)dy=\frac{x}{x^{2}+1}dx[/tex]

[tex]\int tanydy=\int \frac{x}{x^2+1}dx \\ -ln|cos y|=\frac{ln|x^{2}+1|}{2}+C[/tex]

[tex]e^{-ln(cosy)}=e^{\frac{ln(x^{2}+1)}{2}+C}[/tex]

[tex]\frac{1}{cosy}=e^{C}(\sqrt{x^2+1})[/tex]

[tex]cosy=\frac{1}{D\sqrt{x^{2}+1}} \implies y=cos^{-1}\bigg(\frac{1}{D\sqrt{x^2+1}}\bigg)[/tex]

D represents e^C just fyi.

Wondering if this is correct and if my work makes sense.

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