What is the integral of (1-8x)(ln(x)+1) ?

The integral of (1-8x)(ln(x)+1) is xln(x)-4x^2ln(x)-2x^2+C

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integral of (1-8x)(ln(x)+1)

\int (1 - 8 x) (ln (x) + 1) d x

x ln (x) - 4 x^{2} ln (x) - 2 x^{2} + C

Solution steps

\int (1 - 8 x) (ln (x) + 1) d x

Expand

(1 - 8 x) (ln (x) + 1) : ln (x) + 1 - 8 x ln (x) - 8 x

= \int ln (x) + 1 - 8 x ln (x) - 8 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 1 d x - \int 8 x ln (x) d x - \int 8 x d x

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

\int 1 d x = x

\int 8 x ln (x) d x = 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4})

\int 8 x d x = 4 x^{2}

= x ln (x) - x + x - 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) - 4 x^{2}

Simplify

x ln (x) - x + x - 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) - 4 x^{2} : x ln (x) - 4 x^{2} ln (x) - 2 x^{2}

= x ln (x) - 4 x^{2} ln (x) - 2 x^{2}

Add a constant to the solution

= x ln (x) - 4 x^{2} ln (x) - 2 x^{2} + C

6 x^{2} + x y + 6 y^{2} = 13, (1, 1)

y^{^{''}} + 9 y = 5 csc^{2} (3 t)

x \to 0 lim (\frac{1}{2 + 3 ^{\frac{1}{x}}})

What is the integral of (1-8x)(ln(x)+1) ?
The integral of (1-8x)(ln(x)+1) is xln(x)-4x^2ln(x)-2x^2+C