What is the integral of (sqrt(1+x^2))/(x^4) ?

The integral of (sqrt(1+x^2))/(x^4) is -((1+x^2)^{3/2})/(3x^3)+C

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integral of (sqrt(1+x^2))/(x^4)

\int \frac{1 + x ^{2}}{x ^{4}} d x

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

Solution steps

\int \frac{1 + x ^{2}}{x ^{4}} d x

Apply Trigonometric Substitution

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

= \int cot^{4} (u) sec^{3} (u) d u

Rewrite using trig identities

= \int \frac{cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

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

= \int \frac{1}{v ^{4}} d v

Apply the Power Rule

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

= - \frac{1}{3 v ^{3}}

Substitute back

= - \frac{1}{3 sin ^{3} ( arctan ( x ) )}

Simplify

- \frac{1}{3 sin ^{3} ( arctan ( x ) )} : - \frac{( 1 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}}

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

Add a constant to the solution

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

\int ((x - 5)^{2} - x)^{2} d x

\int x^{2} e^{\frac{- x}{2}} d x

\int (x^{3} - sec^{2} (x)) d x

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

\int t sec^{2} (t) d t

What is the integral of (sqrt(1+x^2))/(x^4) ?
The integral of (sqrt(1+x^2))/(x^4) is -((1+x^2)^{3/2})/(3x^3)+C