What is the integral of (x^2)/((9+x^2)^{1/2)} ?

The integral of (x^2)/((9+x^2)^{1/2)} is 9(-1/2 ln((|x+(9+x^2)^{1/2}|)/3)+1/18 x(9+x^2)^{1/2})+C

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integral of (x^2)/((9+x^2)^{1/2)}

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

9 - \frac{1}{2} ln \frac{x + ( 9 + x ^{2} ) ^{\frac{1}{2}}}{3} + \frac{1}{18} x (9 + x^{2})^{\frac{1}{2}} + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply Trigonometric Substitution

: \int 9 tan^{2} (u) sec (u) d u

= \int 9 tan^{2} (u) sec (u) d u

Take the constant out:

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

= 9 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 9 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 9 \cdot \int - sec (u) + sec^{3} (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

= 9 (- \int sec (u) d u + \int sec^{3} (u) d u)

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

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 9 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{3} x)

= 9 (- ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)) + \frac{1}{2} sec (arctan (\frac{1}{3} x)) tan (arctan (\frac{1}{3} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)))

Simplify

9 (- ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)) + \frac{1}{2} sec (arctan (\frac{1}{3} x)) tan (arctan (\frac{1}{3} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))) : 9 - \frac{1}{2} ln \frac{x + ( 9 + x ^{2} ) ^{\frac{1}{2}}}{3} + \frac{1}{18} x (9 + x^{2})^{\frac{1}{2}}

= 9 - \frac{1}{2} ln \frac{x + ( 9 + x ^{2} ) ^{\frac{1}{2}}}{3} + \frac{1}{18} x (9 + x^{2})^{\frac{1}{2}}

Add a constant to the solution

= 9 - \frac{1}{2} ln \frac{x + ( 9 + x ^{2} ) ^{\frac{1}{2}}}{3} + \frac{1}{18} x (9 + x^{2})^{\frac{1}{2}} + C

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What is the integral of (x^2)/((9+x^2)^{1/2)} ?
The integral of (x^2)/((9+x^2)^{1/2)} is 9(-1/2 ln((|x+(9+x^2)^{1/2}|)/3)+1/18 x(9+x^2)^{1/2})+C