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

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

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integral of (x^2)/((x^2+1)^{3/2)}

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{d y}{d x} = - \frac{2}{x}

\int \frac{x}{( x ^{2} + 4 ) ^{3}} d x

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

\frac{\partial}{\partial y} (x^{2} + y^{2})

\int_{0}^{8} \frac{5}{x ^{\frac{2}{3}}} d x

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