What is the integral of 1/((x^2+2x+3)^{3/2)} ?

The integral of 1/((x^2+2x+3)^{3/2)} is (x+1)/(2(x^2+2x+3)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((x^2+2x+3)^{3/2)}

\int \frac{1}{( x ^{2} + 2 x + 3 ) ^{\frac{3}{2}}} d x

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

Solution steps

\int \frac{1}{( x ^{2} + 2 x + 3 ) ^{\frac{3}{2}}} d x

Complete the square

x^{2} + 2 x + 3 : (x + 1)^{2} + 2

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

Apply u-substitution

: \int \frac{1}{( u ^{2} + 2 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( u ^{2} + 2 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{1}{2 sec ( v )} d v

= \int \frac{1}{2 sec ( v )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{sec ( v )} d v

Rewrite using trig identities

= \frac{1}{2} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= \frac{1}{2} sin (v)

Substitute back

= \frac{1}{2} sin (arctan (\frac{1}{2 ^{\frac{1}{2}}} (x + 1)))

Simplify

\frac{1}{2} sin (arctan (\frac{1}{2 ^{\frac{1}{2}}} (x + 1))) : \frac{x + 1}{2 ( x ^{2} + 2 x + 3 ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

\frac{d y}{d x} = 6 y x

x \to 0 lim (\frac{5}{x})

\frac{d y}{d x} + 5 e^{x + y} = 0

\int_{1}^{2} (2 x^{2}) - (2 x + 6) d x

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

What is the integral of 1/((x^2+2x+3)^{3/2)} ?
The integral of 1/((x^2+2x+3)^{3/2)} is (x+1)/(2(x^2+2x+3)^{1/2)}+C