What is the integral of (x^2)/(sqrt(64+x^2)) ?

The integral of (x^2)/(sqrt(64+x^2)) is 64(-1/2 ln((|x+sqrt(64+x^2)|)/8)+1/128 xsqrt(64+x^2))+C

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integral of (x^2)/(sqrt(64+x^2))

\int \frac{x ^{2}}{64 + x ^{2}} d x

64 (- \frac{1}{2} ln (\frac{x + 64 + x ^{2}}{8}) + \frac{1}{128} x 64 + x^{2}) + C

Solution steps

\int \frac{x ^{2}}{64 + x ^{2}} d x

Apply Trigonometric Substitution

: \int 64 tan^{2} (u) sec (u) d u

= \int 64 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

= 64 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 64 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 64 \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

= 64 (- \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) ∣

= 64 (- 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}{8} x)

= 64 (- ln tan (arctan (\frac{1}{8} x)) + sec (arctan (\frac{1}{8} x)) + \frac{1}{2} sec (arctan (\frac{1}{8} x)) tan (arctan (\frac{1}{8} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{8} x)) + sec (arctan (\frac{1}{8} x)))

Simplify

64 (- ln tan (arctan (\frac{1}{8} x)) + sec (arctan (\frac{1}{8} x)) + \frac{1}{2} sec (arctan (\frac{1}{8} x)) tan (arctan (\frac{1}{8} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{8} x)) + sec (arctan (\frac{1}{8} x))) : 64 (- \frac{1}{2} ln (\frac{x + 64 + x ^{2}}{8}) + \frac{1}{128} x 64 + x^{2})

= 64 (- \frac{1}{2} ln (\frac{x + 64 + x ^{2}}{8}) + \frac{1}{128} x 64 + x^{2})

Add a constant to the solution

= 64 (- \frac{1}{2} ln (\frac{x + 64 + x ^{2}}{8}) + \frac{1}{128} x 64 + x^{2}) + C

\frac{d}{d x} (lo g_{e} (2 x))

\int - \frac{3}{x ^{2}} + 11 - 4^{x} d x

x \to - π + lim (csc (x))

(4 x - x^{2})^{3}

x \to - \infty lim (e^{x} + 2)

What is the integral of (x^2)/(sqrt(64+x^2)) ?
The integral of (x^2)/(sqrt(64+x^2)) is 64(-1/2 ln((|x+sqrt(64+x^2)|)/8)+1/128 xsqrt(64+x^2))+C