What is the integral of 1/(xsqrt(25x^2+49)) ?

The integral of 1/(xsqrt(25x^2+49)) is 1/7 ln|tan((arctan(5/7 x))/2)|+C

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integral of 1/(xsqrt(25x^2+49))

\int \frac{1}{x 25 x ^{2} + 49} d x

\frac{1}{7} ln tan (\frac{arctan ( \frac{5}{7} x )}{2}) + C

Solution steps

\int \frac{1}{x 25 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{5}{7} x )}{2})

Add a constant to the solution

= \frac{1}{7} ln tan (\frac{arctan ( \frac{5}{7} x )}{2}) + C

\frac{1}{2 s ^{2} + 6 s + 4}

\frac{d}{d x} (\frac{2 ( x ^{2} - 9 )}{x ^{2} - 4})

n = 0 \sum \infty \frac{n !}{8 ^{n}}

(\frac{sin ( x )}{cos ( x )})^{^{'}}

y^{^{''}} + 6 y^{^{'}} + 11 y = 0

What is the integral of 1/(xsqrt(25x^2+49)) ?
The integral of 1/(xsqrt(25x^2+49)) is 1/7 ln|tan((arctan(5/7 x))/2)|+C