What is the integral of (sqrt(x^2+4))/2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(x^2+4))/2

\int \frac{x ^{2} + 4}{2} d x

ln (\frac{1}{2} x + x^{2} + 4) + \frac{1}{4} x x^{2} + 4 + C

Solution steps

\int \frac{x ^{2} + 4}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x^{2} + 4 d x

Apply Trigonometric Substitution

: \int 4 sec^{3} (u) d u

= \frac{1}{2} \cdot \int 4 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} \cdot 4 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{2} \cdot 4 (\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} \cdot 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{2} x)

= \frac{1}{2} \cdot 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} x ) ) sin ( arctan ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x)))

Simplify

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

= ln (\frac{1}{2} x + x^{2} + 4) + \frac{1}{4} x x^{2} + 4

Add a constant to the solution

= ln (\frac{1}{2} x + x^{2} + 4) + \frac{1}{4} x x^{2} + 4 + C

\int 3 (\frac{sin ( 2 x )}{sin ( x )}) d x

\int \frac{1}{x ( ln ( x ^{6} ) ) ^{9}} d x

\int - \frac{6}{x ^{3}} d x

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

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

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