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

The integral from-1 to 1 of (1/(1+x^2))^2 is (pi+2)/4

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

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

\frac{π + 2}{4}

Decimal Notation

1.28539 \dots

Solution steps

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

Simplify

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

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

= \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

Rewrite using trig identities

= \int_{- \frac{π}{4}}^{\frac{π}{4}} \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

= \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} 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

= \frac{1}{2} (\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d u + \int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 u) d u)

\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d u = \frac{π}{2}

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 u) d u = 1

= \frac{1}{2} (\frac{π}{2} + 1)

Simplify

\frac{1}{2} (\frac{π}{2} + 1) : \frac{π + 2}{4}

= \frac{π + 2}{4}

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

\int 6 x^{5} - 7 x^{4} - 9 x^{2} d x

\int x + c d x

\int_{0}^{1} \int_{0}^{x} (\frac{6}{7} (x^{2} + x \cdot \frac{y}{2})) d y d x

y = \frac{t}{( t - 7 ) ^{2}}

What is the integral from-1 to 1 of (1/(1+x^2))^2 ?
The integral from-1 to 1 of (1/(1+x^2))^2 is (pi+2)/4