What is the integral of sqrt(1+0.16x^2) ?

The integral of sqrt(1+0.16x^2) is 2.5(0.2xsqrt(1+0.16x^2)+0.5ln|0.4x+sqrt(1+0.16x^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+0.16x^2)

\int 1 + 0.16 x^{2} d x

2.5 (0.2 x 1 + 0.16 x^{2} + 0.5 ln 0.4 x + 1 + 0.16 x^{2}) + C

Solution steps

\int 1 + 0.16 x^{2} d x

Apply Trigonometric Substitution

: \int 2.5 sec^{3} (u) d u

= \int 2.5 sec^{3} (u) d u

Take the constant out:

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

= 2.5 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 2.5 (\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) ∣

= 2.5 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (0.4 x)

= 2.5 (\frac{sec ^{2} ( arctan ( 0.4 x ) ) sin ( arctan ( 0.4 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (0.4 x)) + sec (arctan (0.4 x)) ∣)

Simplify

2.5 (\frac{sec ^{2} ( arctan ( 0.4 x ) ) sin ( arctan ( 0.4 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (0.4 x)) + sec (arctan (0.4 x)) ∣) : 2.5 (0.2 x 1 + 0.16 x^{2} + 0.5 ln 0.4 x + 1 + 0.16 x^{2})

= 2.5 (0.2 x 1 + 0.16 x^{2} + 0.5 ln 0.4 x + 1 + 0.16 x^{2})

Add a constant to the solution

= 2.5 (0.2 x 1 + 0.16 x^{2} + 0.5 ln 0.4 x + 1 + 0.16 x^{2}) + C

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

\int \frac{25}{x ^{\frac{1}{2}}} d x

(csc (x) - cot (x))^{- 1}

What is the integral of sqrt(1+0.16x^2) ?
The integral of sqrt(1+0.16x^2) is 2.5(0.2xsqrt(1+0.16x^2)+0.5ln|0.4x+sqrt(1+0.16x^2)|)+C