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

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

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

integral

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

= 2 \cdot \int \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

Rewrite using trig identities

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

Take the constant out:

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

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

\int 1 d u = u

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

= 2 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arctan (x)

= 2 \cdot \frac{1}{2} (arctan (x) + \frac{1}{2} sin (2 arctan (x)))

Simplify

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

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

Add a constant to the solution

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

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

2 + 2 cos^{4} (t)

\frac{( lo g _{10} ( x ) ) \cdot 1}{x}

\frac{x}{x ^{2} - 15}

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

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