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

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

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

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

Solution steps

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

\frac{d}{d x} (e^{x} + 3)

\int_{1}^{9} u + 2 d u

\int \frac{x ^{4} + 3 x ^{2} + 1}{x ^{5} + 5 x ^{3} + 5 x} d x

\int e^{\frac{- x}{2}} d x

cot (x^{5})

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