What is the integral of (100x^4+1)ln(x) ?

The integral of (100x^4+1)ln(x) is 100(1/5 x^5ln(x)-(x^5)/(25))+xln(x)-x+C

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integral of (100x^4+1)ln(x)

\int (100 x^{4} + 1) ln (x) d x

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

Solution steps

\int (100 x^{4} + 1) ln (x) d x

Expand

(100 x^{4} + 1) ln (x) : 100 x^{4} ln (x) + ln (x)

= \int 100 x^{4} ln (x) + 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 100 x^{4} ln (x) d x + \int ln (x) d x

\int 100 x^{4} ln (x) d x = 100 (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25})

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

= 100 (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}) + x ln (x) - x

Add a constant to the solution

= 100 (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}) + x ln (x) - x + C

\frac{d}{d x} (\frac{x ^{2} - 4}{x - 2})

\frac{d}{d x} (ln (x^{3} + 6 x^{2}))

f (x) = x^{2} - 8 x^{2}

\frac{\partial}{\partial x} (\frac{w ^{T} e ^{- w^{T} x}}{( 1 + e ^{- w^{T} x} ) ^{2}})

\frac{1}{( b ^{2} \cdot ( \frac{1 + x}{b ^{2}} ) )}

What is the integral of (100x^4+1)ln(x) ?
The integral of (100x^4+1)ln(x) is 100(1/5 x^5ln(x)-(x^5)/(25))+xln(x)-x+C