What is the integral of 1/(sqrt(x^2+4/5)) ?

The integral of 1/(sqrt(x^2+4/5)) is ln(1/2 |sqrt(5)x+sqrt(4+5x^2)|)+C

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integral of 1/(sqrt(x^2+4/5))

\int \frac{1}{x ^{2} + \frac{4}{5}} d x

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

Solution steps

\int \frac{1}{x ^{2} + \frac{4}{5}} d x

x^{2} + \frac{4}{5} = \frac{4}{5} (\frac{x ^{2}}{\frac{4}{5}} + 1)

= \int \frac{1}{\frac{4}{5} ( \frac{x ^{2}}{\frac{4}{5}} + 1 )} d x

\frac{1}{\frac{4}{5} ( \frac{x ^{2}}{\frac{4}{5}} + 1 )} = \frac{1}{\frac{4}{5}} \frac{1}{\frac{x ^{2}}{\frac{4}{5}} + 1}

= \int \frac{1}{\frac{4}{5}} \cdot \frac{1}{\frac{x ^{2}}{\frac{4}{5}} + 1} d x

Take the constant out:

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

= \frac{1}{\frac{4}{5}} \cdot \int \frac{1}{\frac{x ^{2}}{\frac{4}{5}} + 1} d x

\frac{1}{\frac{4}{5}} = \frac{5}{2}

= \frac{5}{2} \cdot \int \frac{1}{\frac{x ^{2}}{\frac{4}{5}} + 1} d x

Apply Trigonometric Substitution

= \frac{5}{2} \cdot \int \frac{2 sec ( u )}{5} d u

Take the constant out:

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

= \frac{5}{2} \cdot \frac{2}{5} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{5}{2} \cdot \frac{2}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{5}{2} x)

= \frac{5}{2} \cdot \frac{2}{5} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x))

Simplify

\frac{5}{2} \cdot \frac{2}{5} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x)) : ln (\frac{1}{2} 5 x + 4 + 5 x^{2})

= ln (\frac{1}{2} 5 x + 4 + 5 x^{2})

Add a constant to the solution

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

What is the integral of 1/(sqrt(x^2+4/5)) ?
The integral of 1/(sqrt(x^2+4/5)) is ln(1/2 |sqrt(5)x+sqrt(4+5x^2)|)+C