What is the integral of (sqrt(1+x^8))/(x^{13)} ?

The integral of (sqrt(1+x^8))/(x^{13)} is -((1+x^8)^{3/2})/(12x^{12)}+C

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integral of (sqrt(1+x^8))/(x^{13)}

\int \frac{1 + x ^{8}}{x ^{13}} d x

- \frac{( 1 + x ^{8} ) ^{\frac{3}{2}}}{12 x ^{12}} + C

Solution steps

\int \frac{1 + x ^{8}}{x ^{13}} d x

Apply u-substitution

: \int \frac{1 + u}{8 u ^{\frac{5}{2}}} d u

= \int \frac{1 + u}{8 u ^{\frac{5}{2}}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1 + u}{u ^{\frac{5}{2}}} d u

Apply u-substitution

: \int \frac{2 v ^{2}}{( v ^{2} - 1 ) ^{\frac{5}{2}}} d v

= \frac{1}{8} \cdot \int \frac{2 v ^{2}}{( v ^{2} - 1 ) ^{\frac{5}{2}}} d v

Take the constant out:

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

= \frac{1}{8} \cdot 2 \cdot \int \frac{v ^{2}}{( v ^{2} - 1 ) ^{\frac{5}{2}}} d v

Apply Trigonometric Substitution

: \int cot^{4} (w) sec^{3} (w) d w

= \frac{1}{8} \cdot 2 \cdot \int cot^{4} (w) sec^{3} (w) d w

Rewrite using trig identities

= \frac{1}{8} \cdot 2 \cdot \int \frac{cos ( w )}{sin ^{4} ( w )} d w

Apply u-substitution

: \int \frac{1}{t ^{4}} d t

= \frac{1}{8} \cdot 2 \cdot \int \frac{1}{t ^{4}} d t

Apply the Power Rule

: - \frac{1}{3 t ^{3}}

= \frac{1}{8} \cdot 2 (- \frac{1}{3 t ^{3}})

Substitute back

= \frac{1}{8} \cdot 2 (- \frac{1}{3 sin ^{3} ( arcsec ( 1 + x ^{8} ) )})

Simplify

\frac{1}{8} \cdot 2 (- \frac{1}{3 sin ^{3} ( arcsec ( 1 + x ^{8} ) )}) : - \frac{( 1 + x ^{8} ) ^{\frac{3}{2}}}{12 x ^{12}}

= - \frac{( 1 + x ^{8} ) ^{\frac{3}{2}}}{12 x ^{12}}

Add a constant to the solution

= - \frac{( 1 + x ^{8} ) ^{\frac{3}{2}}}{12 x ^{12}} + C

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What is the integral of (sqrt(1+x^8))/(x^{13)} ?
The integral of (sqrt(1+x^8))/(x^{13)} is -((1+x^8)^{3/2})/(12x^{12)}+C