What is the integral of 1/(x^2sqrt(49x^2+16)) ?

The integral of 1/(x^2sqrt(49x^2+16)) is -(sqrt(16+49x^2))/(16x)+C

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integral of 1/(x^2sqrt(49x^2+16))

\int \frac{1}{x ^{2} 49 x ^{2} + 16} d x

- \frac{16 + 49 x ^{2}}{16 x} + C

Solution steps

\int \frac{1}{x ^{2} 49 x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int \frac{7 sec ( u )}{16 tan ^{2} ( u )} d u

= \int \frac{7 sec ( u )}{16 tan ^{2} ( u )} d u

Take the constant out:

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

= \frac{7}{16} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{7}{16} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{7}{16} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{7}{16} (- \frac{1}{v})

Substitute back

= \frac{7}{16} (- \frac{1}{sin ( arctan ( \frac{7}{4} x ) )})

Simplify

\frac{7}{16} (- \frac{1}{sin ( arctan ( \frac{7}{4} x ) )}) : - \frac{16 + 49 x ^{2}}{16 x}

= - \frac{16 + 49 x ^{2}}{16 x}

Add a constant to the solution

= - \frac{16 + 49 x ^{2}}{16 x} + C

\frac{d}{d t} (e^{t a n (π t)})

\frac{x ^{2}}{e ^{x}}

\int arcsin (x) 1 - x d x

r \to 3 lim (\frac{4}{3} π r^{3})

x \to \infty lim (\frac{1 + 3 x + 2 x ^{3}}{x ^{3}})

What is the integral of 1/(x^2sqrt(49x^2+16)) ?
The integral of 1/(x^2sqrt(49x^2+16)) is -(sqrt(16+49x^2))/(16x)+C