What is the integral of (x^2)/(sqrt((1+x^2)^3)) ?

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

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integral of (x^2)/(sqrt((1+x^2)^3))

\int \frac{x ^{2}}{( 1 + x ^{2} ) ^{3}} d x

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

Solution steps

\int \frac{x ^{2}}{( 1 + x ^{2} ) ^{3}} d x

(1 + x^{2})^{3} = (1 + x^{2})^{\frac{3}{2}},

assuming

(1 + x^{2}) \geq 0

= \int \frac{x ^{2}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

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

Simplify

tan^{2} (u) cos (u) : sin (u) tan (u)

= \int sin (u) tan (u) d u

Rewrite using trig identities

= \int \frac{1 - cos ^{2} ( u )}{cos ( u )} d u

Expand

\frac{1 - cos ^{2} ( u )}{cos ( u )} : \frac{1}{cos ( u )} - cos (u)

= \int \frac{1}{cos ( u )} - cos (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 \frac{1}{cos ( u )} d u - \int cos (u) d u

\int \frac{1}{cos ( u )} d u = ln ∣ tan (u) + sec (u) ∣

\int cos (u) d u = sin (u)

= ln ∣ tan (u) + sec (u) ∣ - sin (u)

Substitute back

u = arctan (x)

= ln ∣ tan (arctan (x)) + sec (arctan (x)) ∣ - sin (arctan (x))

Simplify

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

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

Add a constant to the solution

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

\int 7 - 3 x d x

\int \frac{8 x ^{3} - 9 x + 7}{x ^{3} - x ^{2}} d x

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

\int (e^{x} + \frac{1}{5 x}) d x

\int (tan (t) + sin (t)) cos (t) d t

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