What is the integral from 1 to 2 of (9x^2-4x+1)ln(x) ?

The integral from 1 to 2 of (9x^2-4x+1)ln(x) is 18ln(2)-5

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integral from 1 to 2 of (9x^2-4x+1)ln(x)

\int_{1}^{2} (9 x^{2} - 4 x + 1) ln (x) d x

18 ln (2) - 5

Decimal Notation

7.47664 \dots

Solution steps

\int_{1}^{2} (9 x^{2} - 4 x + 1) ln (x) d x

Expand

(9 x^{2} - 4 x + 1) ln (x) : 9 x^{2} ln (x) - 4 x ln (x) + ln (x)

= \int_{1}^{2} 9 x^{2} ln (x) - 4 x 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_{1}^{2} 9 x^{2} ln (x) d x - \int_{1}^{2} 4 x ln (x) d x + \int_{1}^{2} ln (x) d x

\int_{1}^{2} 9 x^{2} ln (x) d x = - 7 + 24 ln (2)

\int_{1}^{2} 4 x ln (x) d x = 8 ln (2) - 3

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

= - 7 + 24 ln (2) - (8 ln (2) - 3) + 2 ln (2) - 1

Simplify

- 7 + 24 ln (2) - (8 ln (2) - 3) + 2 ln (2) - 1 : 18 ln (2) - 5

= 18 ln (2) - 5

\int_{- 1}^{3} 4 x (1 - x^{2}) d x

\int_{3}^{5} \frac{1}{x ^{2}} d x

\int_{\frac{π}{6}}^{\frac{π}{2}} x csc^{2} (x) d x

\int_{- 2}^{0} (4 x^{2} + 4 x) d x

\int_{- 3}^{2} x d x

What is the integral from 1 to 2 of (9x^2-4x+1)ln(x) ?
The integral from 1 to 2 of (9x^2-4x+1)ln(x) is 18ln(2)-5