What is the integral of (x^2)/(sqrt(x^2+1)) ?

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

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integral of (x^2)/(sqrt(x^2+1))

\int \frac{x ^{2}}{x ^{2} + 1} d x

\frac{1}{2} x x^{2} + 1 - \frac{1}{2} ln x + x^{2} + 1 + C

Solution steps

\int \frac{x ^{2}}{x ^{2} + 1} d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Expand

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

= \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

= - \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) ∣

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

Substitute back

u = arctan (x)

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

Simplify

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

= \frac{1}{2} x x^{2} + 1 - \frac{1}{2} ln x + x^{2} + 1

Add a constant to the solution

= \frac{1}{2} x x^{2} + 1 - \frac{1}{2} ln x + x^{2} + 1 + C

(y + 1) (\frac{d y}{d x}) = 2 x

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = - \frac{1}{2}, y^{^{'}} (0) = - \frac{9}{4}

\int \frac{15 e ^{5 y}}{e ^{5 y} + 2} d y

x^{3} + y^{3} = 8 x y, (4, 4)

x \to 3 lim (1 0^{x^{2} - 3 \cdot x})

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