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

The integral from 1-3x to 1 of (u^3)/(1+u^2) is 1/2 (-9x^2+6x-ln(2)+ln|9x^2-6x+2|)

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

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

\frac{1}{2} (- 9 x^{2} + 6 x - ln (2) + ln 9 x^{2} - 6 x + 2)

Solution steps

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

Apply u-substitution

: \int_{9 x^{2} - 6 x + 2}^{2} \frac{v - 1}{2 v} d v

= \int_{9 x^{2} - 6 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_{9 x^{2} - 6 x + 2}^{2} \frac{v - 1}{v} d v

Expand

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

= \frac{1}{2} \cdot \int_{9 x^{2} - 6 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_{9 x^{2} - 6 x + 2}^{2} 1 d v - \int_{9 x^{2} - 6 x + 2}^{2} \frac{1}{v} d v)

\int_{9 x^{2} - 6 x + 2}^{2} 1 d v = - 9 x^{2} + 6 x

\int_{9 x^{2} - 6 x + 2}^{2} \frac{1}{v} d v = ln (2) - ln 9 x^{2} - 6 x + 2

= \frac{1}{2} (- 9 x^{2} + 6 x - (ln (2) - ln 9 x^{2} - 6 x + 2))

Simplify

= \frac{1}{2} (- 9 x^{2} + 6 x - ln (2) + ln 9 x^{2} - 6 x + 2)

\int 16 + 6 x - x^{2} d x

n \to \infty lim (5)

\int - x^{2} + 6 x - 5 d x

cos (2 x), π

f (x) = ln (x^{2} + 3)

What is the integral from 1-3x to 1 of (u^3)/(1+u^2) ?
The integral from 1-3x to 1 of (u^3)/(1+u^2) is 1/2 (-9x^2+6x-ln(2)+ln|9x^2-6x+2|)