What is the integral of (2x-1)ln(17x) ?

The integral of (2x-1)ln(17x) is x^2ln(17x)-xln(17x)-(x^2)/2+x+C

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integral of (2x-1)ln(17x)

\int (2 x - 1) ln (17 x) d x

x^{2} ln (17 x) - x ln (17 x) - \frac{x ^{2}}{2} + x + C

Solution steps

\int (2 x - 1) ln (17 x) d x

Apply Integration By Parts

: ln (17 x) (x^{2} - x) - \int x - 1 d x

= ln (17 x) (x^{2} - x) - \int x - 1 d x

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

= ln (17 x) (x^{2} - x) - (\frac{x ^{2}}{2} - x)

Simplify

ln (17 x) (x^{2} - x) - (\frac{x ^{2}}{2} - x) : x^{2} ln (17 x) - x ln (17 x) - \frac{x ^{2}}{2} + x

= x^{2} ln (17 x) - x ln (17 x) - \frac{x ^{2}}{2} + x

Add a constant to the solution

= x^{2} ln (17 x) - x ln (17 x) - \frac{x ^{2}}{2} + x + C

\frac{2}{x}, 4 x, \frac{4}{x}

5 x

x y^{^{'}} - 4 y = x^{6} e^{x}

x \to \infty lim (\frac{x ^{99}}{e ^{x}})

\int_{1}^{a} (\frac{1}{4 x}) d x

What is the integral of (2x-1)ln(17x) ?
The integral of (2x-1)ln(17x) is x^2ln(17x)-xln(17x)-(x^2)/2+x+C