What is the integral of 1/(sqrt(60x+10x^2)) ?

The integral of 1/(sqrt(60x+10x^2)) is 1/(sqrt(10))ln((|sqrt(x^2+6x)+x+3|)/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(60x+10x^2))

\int \frac{1}{60 x + 10 x ^{2}} d x

\frac{1}{10} ln (\frac{x ^{2} + 6 x + x + 3}{3}) + C

Solution steps

\int \frac{1}{60 x + 10 x ^{2}} d x

Complete the square

60 x + 10 x^{2} : 10 (x + 3)^{2} - 90

= \int \frac{1}{10 ( x + 3 ) ^{2} - 90} d x

Apply u-substitution

: \int \frac{1}{10 u ^{2} - 90} d u

= \int \frac{1}{10 u ^{2} - 90} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{10} d v

= \int \frac{sec ( v )}{10} d v

Take the constant out:

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

= \frac{1}{10} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{10} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{10} ln tan (arcsec (\frac{10}{3 10} (x + 3))) + sec (arcsec (\frac{10}{3 10} (x + 3)))

Simplify

\frac{1}{10} ln tan (arcsec (\frac{10}{3 10} (x + 3))) + sec (arcsec (\frac{10}{3 10} (x + 3))) : \frac{1}{10} ln (\frac{x ^{2} + 6 x + x + 3}{3})

= \frac{1}{10} ln (\frac{x ^{2} + 6 x + x + 3}{3})

Add a constant to the solution

= \frac{1}{10} ln (\frac{x ^{2} + 6 x + x + 3}{3}) + C

\frac{d y}{d x} + y^{2} sin (x) = 0

\int \frac{x ^{2}}{1 + x ^{6}} d x

\int \frac{7}{( ( x - 1 ) x ^{2} - 2 x - 48 )} d x

y^{^{'}} = 3 y + t

\frac{1}{π} sin (π x)

What is the integral of 1/(sqrt(60x+10x^2)) ?
The integral of 1/(sqrt(60x+10x^2)) is 1/(sqrt(10))ln((|sqrt(x^2+6x)+x+3|)/3)+C