What is the integral of 1/(xsqrt(25x^2+16)) ?

The integral of 1/(xsqrt(25x^2+16)) is 1/4 ln|tan((arctan(5/4 x))/2)|+C

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integral of 1/(xsqrt(25x^2+16))

\int \frac{1}{x 25 x ^{2} + 16} d x

\frac{1}{4} ln tan (\frac{arctan ( \frac{5}{4} x )}{2}) + C

Solution steps

\int \frac{1}{x 25 x ^{2} + 16} d x

Apply Trigonometric Substitution

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

= \int \frac{sec ( u )}{4 tan ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

= \frac{1}{4} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{4} ln ∣ v ∣

Substitute back

= \frac{1}{4} ln tan (\frac{arctan ( \frac{5}{4} x )}{2})

Add a constant to the solution

= \frac{1}{4} ln tan (\frac{arctan ( \frac{5}{4} x )}{2}) + C

x + y \frac{d y}{d x} = 0

\frac{d}{d x} (3 e^{2 x} + 2 e^{2 x} x)

f (x) = x^{2} + 7, 16, at x = 3

\int \frac{e ^{- 3 t}}{t} d t

\frac{1}{3} (3 - 5 x)^{3}

What is the integral of 1/(xsqrt(25x^2+16)) ?
The integral of 1/(xsqrt(25x^2+16)) is 1/4 ln|tan((arctan(5/4 x))/2)|+C