What is the integral of (ln(4x))/(x^2) ?

The integral of (ln(4x))/(x^2) is -(ln(4x))/x-1/x+C

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integral of (ln(4x))/(x^2)

\int \frac{ln ( 4 x )}{x ^{2}} d x

- \frac{ln ( 4 x )}{x} - \frac{1}{x} + C

Solution steps

\int \frac{ln ( 4 x )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{ln ( 4 x )}{x} - \int - \frac{1}{x ^{2}} d x

= - \frac{ln ( 4 x )}{x} - \int - \frac{1}{x ^{2}} d x

\int - \frac{1}{x ^{2}} d x = \frac{1}{x}

= - \frac{ln ( 4 x )}{x} - \frac{1}{x}

Add a constant to the solution

= - \frac{ln ( 4 x )}{x} - \frac{1}{x} + C

\frac{\partial}{\partial y} (y ln (y) - e^{- x y})

\int \frac{3 x + 1}{x ^{2} - 10 x + 28} d x

200 (e)^{0.03 x}

y = x^{4} - 3 x^{2} + 3, y = x^{2}

What is the integral of (ln(4x))/(x^2) ?
The integral of (ln(4x))/(x^2) is -(ln(4x))/x-1/x+C