What is the integral of (x^4+2x-9)ln(x) ?

The integral of (x^4+2x-9)ln(x) is 1/5 x^5ln(x)-(x^5}{25}+x^2ln(x)-\frac{x^2)/2-9(xln(x)-x)+C

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integral of (x^4+2x-9)ln(x)

\int (x^{4} + 2 x - 9) ln (x) d x

\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25} + x^{2} ln (x) - \frac{x ^{2}}{2} - 9 (x ln (x) - x) + C

Solution steps

\int (x^{4} + 2 x - 9) ln (x) d x

Expand

(x^{4} + 2 x - 9) ln (x) : x^{4} ln (x) + 2 x ln (x) - 9 ln (x)

= \int x^{4} ln (x) + 2 x ln (x) - 9 ln (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 x^{4} ln (x) d x + \int 2 x ln (x) d x - \int 9 ln (x) d x

\int x^{4} ln (x) d x = \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}

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

\int 9 ln (x) d x = 9 (x ln (x) - x)

= \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25} + x^{2} ln (x) - \frac{x ^{2}}{2} - 9 (x ln (x) - x)

Add a constant to the solution

= \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25} + x^{2} ln (x) - \frac{x ^{2}}{2} - 9 (x ln (x) - x) + C

\frac{3}{x + 1}

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 3

y^{^{'}} = y (3 - y t)

\frac{d}{d x} (3 \cdot sin (x))

2 x^{2} y^{^{''}} + 5 x y^{^{'}} + y = 18 x + 4

What is the integral of (x^4+2x-9)ln(x) ?
The integral of (x^4+2x-9)ln(x) is 1/5 x^5ln(x)-(x^5}{25}+x^2ln(x)-\frac{x^2)/2-9(xln(x)-x)+C