What is the integral of 1/(xsqrt(49+36x^2)) ?

The integral of 1/(xsqrt(49+36x^2)) is 1/7 ln|tan((arctan(6/7 x))/2)|+C

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integral of 1/(xsqrt(49+36x^2))

\int \frac{1}{x 49 + 36 x ^{2}} d x

\frac{1}{7} ln tan (\frac{arctan ( \frac{6}{7} x )}{2}) + C

Solution steps

\int \frac{1}{x 49 + 36 x ^{2}} 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{6}{7} x )}{2})

Add a constant to the solution

= \frac{1}{7} ln tan (\frac{arctan ( \frac{6}{7} x )}{2}) + C

\int \frac{4 x}{( x - 1 ) ( x + 1 ) ^{2}} d x

1 + sin^{2} (x)

\int 1 + sin (x) cos (x) d x

\int x + \frac{2}{sin ( x )} d x

\int \frac{1}{( x ^{2} - x ) ( x ^{2} + 1 )} d x

What is the integral of 1/(xsqrt(49+36x^2)) ?
The integral of 1/(xsqrt(49+36x^2)) is 1/7 ln|tan((arctan(6/7 x))/2)|+C