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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Expand

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

= 2 \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

= 2 (- \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) ∣

= 2 (- 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)

= 2 (- 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

2 (- 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)) ∣) : x x^{2} + 1 - ln x + x^{2} + 1

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

Add a constant to the solution

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

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