What is the integral of (x^2)/(sqrt(x^2+16)) ?

The integral of (x^2)/(sqrt(x^2+16)) is 16(-1/2 ln((|x+sqrt(16+x^2)|)/4)+1/32 xsqrt(16+x^2))+C

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integral of (x^2)/(sqrt(x^2+16))

\int \frac{x ^{2}}{x ^{2} + 16} d x

16 (- \frac{1}{2} ln (\frac{x + 16 + x ^{2}}{4}) + \frac{1}{32} x 16 + x^{2}) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int 16 tan^{2} (u) sec (u) d u

= \int 16 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

= 16 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 16 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 16 \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

= 16 (- \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) ∣

= 16 (- 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}{4} x)

= 16 (- ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) + \frac{1}{2} sec (arctan (\frac{1}{4} x)) tan (arctan (\frac{1}{4} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)))

Simplify

16 (- ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) + \frac{1}{2} sec (arctan (\frac{1}{4} x)) tan (arctan (\frac{1}{4} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))) : 16 (- \frac{1}{2} ln (\frac{x + 16 + x ^{2}}{4}) + \frac{1}{32} x 16 + x^{2})

= 16 (- \frac{1}{2} ln (\frac{x + 16 + x ^{2}}{4}) + \frac{1}{32} x 16 + x^{2})

Add a constant to the solution

= 16 (- \frac{1}{2} ln (\frac{x + 16 + x ^{2}}{4}) + \frac{1}{32} x 16 + x^{2}) + C

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What is the integral of (x^2)/(sqrt(x^2+16)) ?
The integral of (x^2)/(sqrt(x^2+16)) is 16(-1/2 ln((|x+sqrt(16+x^2)|)/4)+1/32 xsqrt(16+x^2))+C