What is the integral of 1/(sqrt((5-4x-x^2)^3)) ?

The integral of 1/(sqrt((5-4x-x^2)^3)) is (x+2)/(9sqrt(-x^2-4x+5))+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((5-4x-x^2)^3))

\int \frac{1}{( 5 - 4 x - x ^{2} ) ^{3}} d x

\frac{x + 2}{9 - x ^{2} - 4 x + 5} + C

Solution steps

\int \frac{1}{( 5 - 4 x - x ^{2} ) ^{3}} d x

(5 - 4 x - x^{2})^{3} = (5 - 4 x - x^{2})^{\frac{3}{2}},

assuming

(5 - 4 x - x^{2}) \geq 0

= \int \frac{1}{( 5 - 4 x - x ^{2} ) ^{\frac{3}{2}}} d x

Complete the square

5 - 4 x - x^{2} : - (x + 2)^{2} + 9

= \int \frac{1}{( - ( x + 2 ) ^{2} + 9 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{1}{9 cos ^{2} ( v )} d v

= \int \frac{1}{9 cos ^{2} ( v )} d v

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{cos ^{2} ( v )} d v

Use the common integral:

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

= \frac{1}{9} tan (v)

Substitute back

= \frac{1}{9} tan (arcsin (\frac{1}{3} (x + 2)))

Simplify

\frac{1}{9} tan (arcsin (\frac{1}{3} (x + 2))) : \frac{x + 2}{9 - x ^{2} - 4 x + 5}

= \frac{x + 2}{9 - x ^{2} - 4 x + 5}

Add a constant to the solution

= \frac{x + 2}{9 - x ^{2} - 4 x + 5} + C

\int \frac{x}{x ^{2} + x - 12} d x

\int e^{1 - x} cos (2 x) d x

\int x^{3} ln (e^{x}) d x

\int \frac{3 y ^{4} + 12 y ^{2} - 1}{y ^{3} + 4 y} d y

\int + x e - x^{2} d x

What is the integral of 1/(sqrt((5-4x-x^2)^3)) ?
The integral of 1/(sqrt((5-4x-x^2)^3)) is (x+2)/(9sqrt(-x^2-4x+5))+C