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

The integral of (2x)/((x^2+1)^2-x^2) is arctan(x^2)+C

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

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

arctan (x^{2}) + C

Solution steps

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

Expand

(x^{2} + 1)^{2} - x^{2} : x^{4} + x^{2} + 1

= 2 \cdot \int \frac{x}{x ^{4} + x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{4 u ( u + u + 1 )} d u

= 2 \cdot \int \frac{1}{4 u ( u + u + 1 )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + u + 1 )} d u

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= 2 \cdot \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{u + 1} d u

Simplify

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

= 2 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 1 )} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 1} d v

= 2 \cdot \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 1} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{4} \cdot 2 arctan (v)

Substitute back

= 2 \cdot \frac{1}{4} \cdot 2 arctan (x^{4})

Simplify

2 \cdot \frac{1}{4} \cdot 2 arctan (x^{4}) : arctan (x^{2})

= arctan (x^{2})

Add a constant to the solution

= arctan (x^{2}) + C

\int 7 tan^{3} (θ) d θ

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 3 x - y}

y^{3} - (18 x + 2) + 3 x y^{2} y^{^{'}} = 0

y \to - 2 lim (\frac{y ^{3} + 2}{y + 2})

\frac{1}{y ^{2}}

What is the integral of (2x)/((x^2+1)^2-x^2) ?
The integral of (2x)/((x^2+1)^2-x^2) is arctan(x^2)+C