What is the integral of 7/(x^2sqrt(x^2+36)) ?

The integral of 7/(x^2sqrt(x^2+36)) is -(7sqrt(36+x^2))/(36x)+C

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integral of 7/(x^2sqrt(x^2+36))

\int \frac{7}{x ^{2} x ^{2} + 36} d x

- \frac{7 36 + x ^{2}}{36 x} + C

Solution steps

\int \frac{7}{x ^{2} x ^{2} + 36} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{x ^{2} x ^{2} + 36} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{36 tan ^{2} ( u )} d u

= 7 \cdot \int \frac{sec ( u )}{36 tan ^{2} ( u )} d u

Take the constant out:

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

= 7 \cdot \frac{1}{36} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 7 \cdot \frac{1}{36} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 7 \cdot \frac{1}{36} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 7 \cdot \frac{1}{36} (- \frac{1}{v})

Substitute back

= 7 \cdot \frac{1}{36} (- \frac{1}{sin ( arctan ( \frac{1}{6} x ) )})

Simplify

7 \cdot \frac{1}{36} (- \frac{1}{sin ( arctan ( \frac{1}{6} x ) )}) : - \frac{7 36 + x ^{2}}{36 x}

= - \frac{7 36 + x ^{2}}{36 x}

Add a constant to the solution

= - \frac{7 36 + x ^{2}}{36 x} + C

\int \frac{1}{x} e^{x} d x

\frac{d}{d x} (\frac{x}{3 x - 4 \cdot ln ( x )})

f (x) = x^{2} 16 - x^{2}

\frac{\partial}{\partial x} (x \cdot e^{x^{2} y})

f (x) = 3 x^{2} cos (x) + 7 x

What is the integral of 7/(x^2sqrt(x^2+36)) ?
The integral of 7/(x^2sqrt(x^2+36)) is -(7sqrt(36+x^2))/(36x)+C