What is the solution for dy=(9x+4y+1)^2dx ?

The solution for dy=(9x+4y+1)^2dx is y=(3tan(6x+c_{1}))/8-(9x)/4-1/4

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dy=(9x+4y+1)^2dx

d y = (9 x + 4 y + 1)^{2} d x

y = \frac{3 tan ( 6 x + c _{1} )}{8} - \frac{9 x}{4} - \frac{1}{4}

Solution steps

d y = (9 x + 4 y + 1)^{2} d x

Solve using substitution:

y = \frac{3 tan ( 6 x + c _{1} )}{8} - \frac{9 x}{4} - \frac{1}{4}

y = \frac{3 tan ( 6 x + c _{1} )}{8} - \frac{9 x}{4} - \frac{1}{4}

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What is the solution for dy=(9x+4y+1)^2dx ?
The solution for dy=(9x+4y+1)^2dx is y=(3tan(6x+c_{1}))/8-(9x)/4-1/4