What is the integral of (x^{11})/((x^6+1)^{100)} ?

The integral of (x^{11})/((x^6+1)^{100)} is (-99x^6-1)/(58212(x^6+1)^{99)}+C

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integral of (x^{11})/((x^6+1)^{100)}

\int \frac{x ^{11}}{( x ^{6} + 1 ) ^{100}} d x

\frac{- 99 x ^{6} - 1}{58212 ( x ^{6} + 1 ) ^{99}} + C

Solution steps

\int \frac{x ^{11}}{( x ^{6} + 1 ) ^{100}} d x

Apply u-substitution

= \int \frac{u - 1}{6 u ^{100}} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{u - 1}{u ^{100}} d u

Expand

\frac{u - 1}{u ^{100}} : \frac{1}{u ^{99}} - \frac{1}{u ^{100}}

= \frac{1}{6} \cdot \int \frac{1}{u ^{99}} - \frac{1}{u ^{100}} 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

= \frac{1}{6} (\int \frac{1}{u ^{99}} d u - \int \frac{1}{u ^{100}} d u)

\int \frac{1}{u ^{99}} d u = - \frac{1}{98 u ^{98}}

\int \frac{1}{u ^{100}} d u = - \frac{1}{99 u ^{99}}

= \frac{1}{6} (- \frac{1}{98 u ^{98}} - (- \frac{1}{99 u ^{99}}))

Substitute back

u = x^{6} + 1

= \frac{1}{6} (- \frac{1}{98 ( x ^{6} + 1 ) ^{98}} - (- \frac{1}{99 ( x ^{6} + 1 ) ^{99}}))

Simplify

\frac{1}{6} (- \frac{1}{98 ( x ^{6} + 1 ) ^{98}} - (- \frac{1}{99 ( x ^{6} + 1 ) ^{99}})) : \frac{- 99 x ^{6} - 1}{58212 ( x ^{6} + 1 ) ^{99}}

= \frac{- 99 x ^{6} - 1}{58212 ( x ^{6} + 1 ) ^{99}}

Add a constant to the solution

= \frac{- 99 x ^{6} - 1}{58212 ( x ^{6} + 1 ) ^{99}} + C

What is the integral of (x^{11})/((x^6+1)^{100)} ?
The integral of (x^{11})/((x^6+1)^{100)} is (-99x^6-1)/(58212(x^6+1)^{99)}+C