What is the integral of 7/(x^2sqrt(x^2+4)) ?

The integral of 7/(x^2sqrt(x^2+4)) is -(7sqrt(4+x^2))/(4x)+C

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integral of 7/(x^2sqrt(x^2+4))

\int \frac{7}{x ^{2} x ^{2} + 4} d x

- \frac{7 4 + x ^{2}}{4 x} + C

Solution steps

\int \frac{7}{x ^{2} x ^{2} + 4} 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} + 4} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

= 7 \cdot \int \frac{sec ( u )}{4 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}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 7 \cdot \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 7 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 7 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 7 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )})

Simplify

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

= - \frac{7 4 + x ^{2}}{4 x}

Add a constant to the solution

= - \frac{7 4 + x ^{2}}{4 x} + C

x \to 0 + lim (\frac{x}{14 x ^{4} - 3 x})

\int - x^{7} e^{x^{8} - 7} d x

3 x^{4}

y^{^{'}} - 4 tan (2 x) y = sin (2 x)

\int_{0}^{1} \frac{24}{x ^{2} - 36} d x

What is the integral of 7/(x^2sqrt(x^2+4)) ?
The integral of 7/(x^2sqrt(x^2+4)) is -(7sqrt(4+x^2))/(4x)+C