What is the integral of (x^7)/(1+x^4) ?

The integral of (x^7)/(1+x^4) is 1/4 (x^4+1-ln|x^4+1|)+C

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integral of (x^7)/(1+x^4)

\int \frac{x ^{7}}{1 + x ^{4}} d x

\frac{1}{4} (x^{4} + 1 - ln x^{4} + 1) + C

Solution steps

\int \frac{x ^{7}}{1 + x ^{4}} d x

Apply u-substitution

= \int \frac{u ^{\frac{1}{7}}}{7 ( u ^{\frac{4}{7}} + 1 )} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{u ^{\frac{1}{7}}}{u ^{\frac{4}{7}} + 1} d u

Apply u-substitution

= \frac{1}{7} \cdot \int \frac{7 ( v - 1 )}{4 v} d v

Take the constant out:

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

= \frac{1}{7} \cdot \frac{7}{4} \cdot \int \frac{v - 1}{v} d v

Expand

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

= \frac{1}{7} \cdot \frac{7}{4} \cdot \int 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}{7} \cdot \frac{7}{4} (\int 1 d v - \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{7} \cdot \frac{7}{4} (v - ln ∣ v ∣)

Substitute back

= \frac{1}{7} \cdot \frac{7}{4} ((x^{7})^{\frac{4}{7}} + 1 - ln (x^{7})^{\frac{4}{7}} + 1)

Simplify

\frac{1}{7} \cdot \frac{7}{4} ((x^{7})^{\frac{4}{7}} + 1 - ln (x^{7})^{\frac{4}{7}} + 1) : \frac{1}{4} (x^{4} + 1 - ln x^{4} + 1)

= \frac{1}{4} (x^{4} + 1 - ln x^{4} + 1)

Add a constant to the solution

= \frac{1}{4} (x^{4} + 1 - ln x^{4} + 1) + C

What is the integral of (x^7)/(1+x^4) ?
The integral of (x^7)/(1+x^4) is 1/4 (x^4+1-ln|x^4+1|)+C