What is the integral of (x^2)/(sqrt(x^2+8)) ?

The integral of (x^2)/(sqrt(x^2+8)) is 8(-1/4 (2ln|x+sqrt(8+x^2)|-3ln(2))+1/16 xsqrt(8+x^2))+C

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integral of (x^2)/(sqrt(x^2+8))

\int \frac{x ^{2}}{x ^{2} + 8} d x

8 (- \frac{1}{4} (2 ln x + 8 + x^{2} - 3 ln (2)) + \frac{1}{16} x 8 + x^{2}) + C

Solution steps

\int \frac{x ^{2}}{x ^{2} + 8} d x

Apply Trigonometric Substitution

: \int 8 tan^{2} (u) sec (u) d u

= \int 8 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

= 8 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 8 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 8 \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

= 8 (- \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) ∣

= 8 (- 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}{2 2} x)

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

Simplify

8 (- ln tan (arctan (\frac{1}{2 2} x)) + sec (arctan (\frac{1}{2 2} x)) + \frac{1}{2} sec (arctan (\frac{1}{2 2} x)) tan (arctan (\frac{1}{2 2} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{2 2} x)) + sec (arctan (\frac{1}{2 2} x))) : 8 (- \frac{1}{4} (2 ln x + 8 + x^{2} - 3 ln (2)) + \frac{1}{16} x 8 + x^{2})

= 8 (- \frac{1}{4} (2 ln x + 8 + x^{2} - 3 ln (2)) + \frac{1}{16} x 8 + x^{2})

Add a constant to the solution

= 8 (- \frac{1}{4} (2 ln x + 8 + x^{2} - 3 ln (2)) + \frac{1}{16} x 8 + x^{2}) + C

\int s 9^{s} d s

\frac{d}{d x} (6 sin^{2} (2 x) cos (2 x) + 2 x)

x^{2} y^{^{''}} + x y^{^{'}} - y = x

x^{2} y^{^{'}} + 3 x y = e^{- x}

\int 2 x^{2} d x

What is the integral of (x^2)/(sqrt(x^2+8)) ?
The integral of (x^2)/(sqrt(x^2+8)) is 8(-1/4 (2ln|x+sqrt(8+x^2)|-3ln(2))+1/16 xsqrt(8+x^2))+C