What is the solution for y^'=(3x+1)^4(4x+1)^{-2} ?

The solution for y^'=(3x+1)^4(4x+1)^{-2} is y=integral of ((3x+1)^4)/((4x+1)^2)

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y^'=(3x+1)^4(4x+1)^{-2}

y^{^{'}} = (3 x + 1)^{4} (4 x + 1)^{- 2}

y = \int \frac{( 3 x + 1 ) ^{4}}{( 4 x + 1 ) ^{2}} d x

Solution steps

y^{'} (x) = (3 x + 1)^{4} (4 x + 1)^{- 2}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int (3 x + 1)^{4} (4 x + 1)^{- 2} d x

y = \int (3 x + 1)^{4} (4 x + 1)^{- 2} d x

Simplify

y = \int \frac{( 3 x + 1 ) ^{4}}{( 4 x + 1 ) ^{2}} d x

\int_{- 1}^{1} e^{x} cos (nπ x) d x

\frac{d}{d t} (a sin (3 t) + b cos (3 t))

n = 1 \sum \infty \frac{1}{5 ^{\frac{1}{n}}}

x \to - 3 lim (\frac{x + 3}{x ^{2} - 9})

f (x) = 2 x e^{x}

What is the solution for y^'=(3x+1)^4(4x+1)^{-2} ?
The solution for y^'=(3x+1)^4(4x+1)^{-2} is y=integral of ((3x+1)^4)/((4x+1)^2)