What is the integral of (6x^2)/((x^6+1)(x^3+1)) ?

The integral of (6x^2)/((x^6+1)(x^3+1)) is -1/2 ln|x^6+1|+arctan(x^3)+ln|x^3+1|+C

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integral of (6x^2)/((x^6+1)(x^3+1))

\int \frac{6 x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

- \frac{1}{2} ln x^{6} + 1 + arctan (x^{3}) + ln x^{3} + 1 + C

Solution steps

\int \frac{6 x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

Take the constant out:

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

= 6 \cdot \int \frac{x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

Apply u-substitution

: \int \frac{1}{3 ( u ^{2} + 1 ) ( u + 1 )} d u

= 6 \cdot \int \frac{1}{3 ( u ^{2} + 1 ) ( u + 1 )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{( u ^{2} + 1 ) ( u + 1 )} d u

Take the partial fraction of

\frac{1}{( u ^{2} + 1 ) ( u + 1 )} : \frac{- u + 1}{2 ( u ^{2} + 1 )} + \frac{1}{2 ( u + 1 )}

= 6 \cdot \frac{1}{3} \cdot \int \frac{- u + 1}{2 ( u ^{2} + 1 )} + \frac{1}{2 ( u + 1 )} d u

Apply the Sum Rule:

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

= 6 \cdot \frac{1}{3} (\int \frac{- u + 1}{2 ( u ^{2} + 1 )} d u + \int \frac{1}{2 ( u + 1 )} d u)

\int \frac{- u + 1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 + arctan (u))

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

= 6 \cdot \frac{1}{3} (\frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 + arctan (u)) + \frac{1}{2} ln ∣ u + 1 ∣)

Substitute back

u = x^{3}

= 6 \cdot \frac{1}{3} (\frac{1}{2} (- \frac{1}{2} ln (x^{3})^{2} + 1 + arctan (x^{3})) + \frac{1}{2} ln x^{3} + 1)

Simplify

6 \cdot \frac{1}{3} (\frac{1}{2} (- \frac{1}{2} ln (x^{3})^{2} + 1 + arctan (x^{3})) + \frac{1}{2} ln x^{3} + 1) : - \frac{1}{2} ln x^{6} + 1 + arctan (x^{3}) + ln x^{3} + 1

= - \frac{1}{2} ln x^{6} + 1 + arctan (x^{3}) + ln x^{3} + 1

Add a constant to the solution

= - \frac{1}{2} ln x^{6} + 1 + arctan (x^{3}) + ln x^{3} + 1 + C

x \to 0 lim (sin (9 x))

f (x) = 2 x^{7}

\frac{d ^{10}}{d x ^{10}} (\frac{x}{e ^{x}})

y^{^{''}} + y = 0

What is the integral of (6x^2)/((x^6+1)(x^3+1)) ?
The integral of (6x^2)/((x^6+1)(x^3+1)) is -1/2 ln|x^6+1|+arctan(x^3)+ln|x^3+1|+C