What is the integral of (4+x^2)^{1/2} ?

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

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integral of (4+x^2)^{1/2}

\int (4 + x^{2})^{\frac{1}{2}} d x

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

Solution steps

\int (4 + x^{2})^{\frac{1}{2}} d x

Apply Trigonometric Substitution

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

= \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

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

Apply Integral Reduction

= 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) ∣

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

Substitute back

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

= 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

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))) : 4 \frac{1}{8} x (x^{2} + 4)^{\frac{1}{2}} + \frac{1}{2} ln \frac{x + ( 4 + x ^{2} ) ^{\frac{1}{2}}}{2}

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

Add a constant to the solution

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

\int sec^{2} (x) (tan (x) + 1) d x

\int \frac{1}{A} d A

\int \frac{2 x}{( x ^{2} + 5 ) ^{4}} d x

\int \frac{4 x - 3}{( x ^{2} + 4 x + 20 )} d x

\int 4 x [x^{2} + 5]^{3} d x

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