What is the integral of (x^2)/((6-x^2)^{3/2)} ?

The integral of (x^2)/((6-x^2)^{3/2)} is -arcsin(1/(6^{1/2)}x)+x/((6-x^2)^{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 (x^2)/((6-x^2)^{3/2)}

\int \frac{x ^{2}}{( 6 - x ^{2} ) ^{\frac{3}{2}}} d x

- arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + \frac{x}{( 6 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x ^{2}}{( 6 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int tan^{2} (u) d u

= \int tan^{2} (u) d u

Rewrite using trig identities

= \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d u + \int sec^{2} (u) d u

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= - u + tan (u)

Substitute back

u = arcsin (\frac{1}{6 ^{\frac{1}{2}}} x)

= - arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + tan (arcsin (\frac{1}{6 ^{\frac{1}{2}}} x))

Simplify

- arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + tan (arcsin (\frac{1}{6 ^{\frac{1}{2}}} x)) : - arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + \frac{x}{( 6 - x ^{2} ) ^{\frac{1}{2}}}

= - arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + \frac{x}{( 6 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + \frac{x}{( 6 - x ^{2} ) ^{\frac{1}{2}}} + C

\int 8 x^{3} - 12 x^{2} + 4 x d x

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

\int e^{8 x - 3} d x

\int (2 x + 1) 2 d x

\int 6 x e^{2 x^{2}} d x

What is the integral of (x^2)/((6-x^2)^{3/2)} ?
The integral of (x^2)/((6-x^2)^{3/2)} is -arcsin(1/(6^{1/2)}x)+x/((6-x^2)^{1/2)}+C