What is the integral from 1-9x to 1 of (u^3)/(1+u^2) ?

The integral from 1-9x to 1 of (u^3)/(1+u^2) is 1/2 (-81x^2+18x-ln(2)+ln|81x^2-18x+2|)

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integral from 1-9x to 1 of (u^3)/(1+u^2)

\int_{1 - 9 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

\frac{1}{2} (- 81 x^{2} + 18 x - ln (2) + ln 81 x^{2} - 18 x + 2)

Solution steps

\int_{1 - 9 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

Apply u-substitution

: \int_{81 x^{2} - 18 x + 2}^{2} \frac{v - 1}{2 v} d v

= \int_{81 x^{2} - 18 x + 2}^{2} \frac{v - 1}{2 v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{81 x^{2} - 18 x + 2}^{2} \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \frac{1}{2} \cdot \int_{81 x^{2} - 18 x + 2}^{2} 1 - \frac{1}{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{81 x^{2} - 18 x + 2}^{2} 1 d v - \int_{81 x^{2} - 18 x + 2}^{2} \frac{1}{v} d v)

\int_{81 x^{2} - 18 x + 2}^{2} 1 d v = - 81 x^{2} + 18 x

\int_{81 x^{2} - 18 x + 2}^{2} \frac{1}{v} d v = ln (2) - ln 81 x^{2} - 18 x + 2

= \frac{1}{2} (- 81 x^{2} + 18 x - (ln (2) - ln 81 x^{2} - 18 x + 2))

Simplify

= \frac{1}{2} (- 81 x^{2} + 18 x - ln (2) + ln 81 x^{2} - 18 x + 2)

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What is the integral from 1-9x to 1 of (u^3)/(1+u^2) ?
The integral from 1-9x to 1 of (u^3)/(1+u^2) is 1/2 (-81x^2+18x-ln(2)+ln|81x^2-18x+2|)