What is the integral of 1/(x^2sqrt(7+x^2)) ?

The integral of 1/(x^2sqrt(7+x^2)) is -(sqrt(7+x^2))/(7x)+C

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integral of 1/(x^2sqrt(7+x^2))

\int \frac{1}{x ^{2} 7 + x ^{2}} d x

- \frac{7 + x ^{2}}{7 x} + C

Solution steps

\int \frac{1}{x ^{2} 7 + x ^{2}} d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \frac{1}{7} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

= - \frac{7 + x ^{2}}{7 x}

Add a constant to the solution

= - \frac{7 + x ^{2}}{7 x} + C

f (x) = e^{7 x} cos (10 x)

\frac{d y}{d x} = x e^{x^{2}}

\int_{- 1}^{1} x^{2} (1.5 x^{2}) d x

a \cdot t (8 - a) + 8 a

\int (x ln (x^{2} + 1) + (x^{2} + 1) y) d x

What is the integral of 1/(x^2sqrt(7+x^2)) ?
The integral of 1/(x^2sqrt(7+x^2)) is -(sqrt(7+x^2))/(7x)+C