What is the integral of 1/(x^2sqrt(x^2+49)) ?

The integral of 1/(x^2sqrt(x^2+49)) is -(sqrt(49+x^2))/(49x)+C

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integral of 1/(x^2sqrt(x^2+49))

\int \frac{1}{x ^{2} x ^{2} + 49} d x

- \frac{49 + x ^{2}}{49 x} + C

Solution steps

\int \frac{1}{x ^{2} x ^{2} + 49} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{49 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{49 tan ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{49} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{49} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{49} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{49} (- \frac{1}{v})

Substitute back

= \frac{1}{49} (- \frac{1}{sin ( arctan ( \frac{1}{7} x ) )})

Simplify

\frac{1}{49} (- \frac{1}{sin ( arctan ( \frac{1}{7} x ) )}) : - \frac{49 + x ^{2}}{49 x}

= - \frac{49 + x ^{2}}{49 x}

Add a constant to the solution

= - \frac{49 + x ^{2}}{49 x} + C

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

5 x^{2} - 2 x - 1

\int \frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9} d x

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x} ln (x)

f (x) = \frac{1}{7 - x}

What is the integral of 1/(x^2sqrt(x^2+49)) ?
The integral of 1/(x^2sqrt(x^2+49)) is -(sqrt(49+x^2))/(49x)+C