What is the integral of 1/(xsqrt(4x^2+49)) ?

The integral of 1/(xsqrt(4x^2+49)) is 1/7 ln|tan((arctan(2/7 x))/2)|+C

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integral of 1/(xsqrt(4x^2+49))

\int \frac{1}{x 4 x ^{2} + 49} d x

\frac{1}{7} ln tan (\frac{arctan ( \frac{2}{7} x )}{2}) + C

Solution steps

\int \frac{1}{x 4 x ^{2} + 49} d x

Apply Trigonometric Substitution

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{7} ln ∣ v ∣

Substitute back

= \frac{1}{7} ln tan (\frac{arctan ( \frac{2}{7} x )}{2})

Add a constant to the solution

= \frac{1}{7} ln tan (\frac{arctan ( \frac{2}{7} x )}{2}) + C

(7, - 2), (- 7, 2)

\frac{d}{d x} (- 27 e^{- 3 x})

n = 0 \sum \infty 5 (4 x^{2})^{n}

\frac{d}{d x} (e^{- 0.01 x})

What is the integral of 1/(xsqrt(4x^2+49)) ?
The integral of 1/(xsqrt(4x^2+49)) is 1/7 ln|tan((arctan(2/7 x))/2)|+C