What is the integral of (4x-1)*ln(x) ?

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

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integral of (4x-1)*ln(x)

\int (4 x - 1) \cdot ln (x) d x

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

Solution steps

\int (4 x - 1) ln (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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