What is the integral of sqrt(1+8x^2) ?

The integral of sqrt(1+8x^2) is 1/2 xsqrt(8x^2+1)+1/(4sqrt(2))ln|2sqrt(2)x+sqrt(1+8x^2)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(1+8x^2)

\int 1 + 8 x^{2} d x

\frac{1}{2} x 8 x^{2} + 1 + \frac{1}{4 2} ln 22 x + 1 + 8 x^{2} + C

Solution steps

\int 1 + 8 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{1}{2 2} sec^{3} (u) d u

= \int \frac{1}{2 2} sec^{3} (u) d u

Take the constant out:

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

= \frac{1}{2 2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{2 2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{2 2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (22 x)

= \frac{1}{2 2} (\frac{sec ^{2} ( arctan ( 2 2 x ) ) sin ( arctan ( 2 2 x ) )}{2} + \frac{1}{2} ln tan (arctan (22 x)) + sec (arctan (22 x)))

Simplify

\frac{1}{2 2} (\frac{sec ^{2} ( arctan ( 2 2 x ) ) sin ( arctan ( 2 2 x ) )}{2} + \frac{1}{2} ln tan (arctan (22 x)) + sec (arctan (22 x))) : \frac{1}{2} x 8 x^{2} + 1 + \frac{1}{4 2} ln 22 x + 1 + 8 x^{2}

= \frac{1}{2} x 8 x^{2} + 1 + \frac{1}{4 2} ln 22 x + 1 + 8 x^{2}

Add a constant to the solution

= \frac{1}{2} x 8 x^{2} + 1 + \frac{1}{4 2} ln 22 x + 1 + 8 x^{2} + C

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

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

\int \frac{y - 8}{y ^{2} - y - 6} d y

\int cot (x) sin (x) d x

What is the integral of sqrt(1+8x^2) ?
The integral of sqrt(1+8x^2) is 1/2 xsqrt(8x^2+1)+1/(4sqrt(2))ln|2sqrt(2)x+sqrt(1+8x^2)|+C