What is the integral of (ln(x)+4x) ?

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

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integral of (ln(x)+4x)

\int (ln (x) + 4 x) d x

x ln (x) - x + 2 x^{2} + C

Solution steps

\int ln (x) + 4 x d x

Apply the Sum Rule:

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

= \int ln (x) d x + \int 4 x d x

\int ln (x) d x = x ln (x) - x

\int 4 x d x = 2 x^{2}

= x ln (x) - x + 2 x^{2}

Add a constant to the solution

= x ln (x) - x + 2 x^{2} + C

\int (x + 3) 8 - x d x

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\frac{d}{d x} (cos (x) - cos^{2} (x))

\frac{d}{d x} (x^{3} + 3 x^{2})

\int_{0}^{1} (x^{e} + e^{x}) d x

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