What is the integral of (16)/((64x^2+1)^2) ?

The integral of (16)/((64x^2+1)^2) is arctan(8x)+1/2 sin(2arctan(8x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (16)/((64x^2+1)^2)

\int \frac{16}{( 64 x ^{2} + 1 ) ^{2}} d x

arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)) + C

Solution steps

\int \frac{16}{( 64 x ^{2} + 1 ) ^{2}} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{( 64 x ^{2} + 1 ) ^{2}} d x

Apply Integral Substitution

= 16 \cdot \int \frac{1}{8 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

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

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= 16 \cdot \frac{1}{8} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

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

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)))

Simplify

16 \cdot \frac{1}{8} \cdot \frac{1}{2} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))) : arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))

= arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))

Add a constant to the solution

= arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)) + C

f (x) = (x + 3)^{3}, [- 1, 6]

\int_{0}^{3} x (3 - x) d x

\int 2 x^{3} (t^{4} - 3)^{5} d x

\frac{3}{( s + 1 ) ( s + 2 )}

\int \frac{14 x - 11}{( 2 x - 3 ) ( x + 1 )} d x

What is the integral of (16)/((64x^2+1)^2) ?
The integral of (16)/((64x^2+1)^2) is arctan(8x)+1/2 sin(2arctan(8x))+C