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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{π + 2}{4}

Decimal Notation

1.28539 \dots

Solution steps

\int_{- 1}^{1} \frac{1}{( x ^{2} + 1 ) ^{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_{1}^{4} \frac{- 3}{( 2 x + 1 ) ^{2}} d x

\int_{0}^{x} t^{6} 1 + t^{7} d t

\int_{1}^{5} \frac{ln ( s )}{s ^{5}} d s

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

\int_{- a}^{a} (a^{2} - t^{2}) d t

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