What is the integral of 5sqrt(x^2+4) ?

The integral of 5sqrt(x^2+4) is 20(1/8 xsqrt(4+x^2)+1/2 ln((|x+sqrt(4+x^2)|)/2))+C

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integral of 5sqrt(x^2+4)

\int 5 x^{2} + 4 d x

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

Solution steps

\int 5 x^{2} + 4 d x

Take the constant out:

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

= 5 \cdot \int x^{2} + 4 d x

Apply Trigonometric Substitution

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

= 5 \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

= 5 \cdot 4 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 5 \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) ∣

= 5 \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)

= 5 \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

5 \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))) : 20 (\frac{1}{8} x 4 + x^{2} + \frac{1}{2} ln (\frac{x + 4 + x ^{2}}{2}))

= 20 (\frac{1}{8} x 4 + x^{2} + \frac{1}{2} ln (\frac{x + 4 + x ^{2}}{2}))

Add a constant to the solution

= 20 (\frac{1}{8} x 4 + x^{2} + \frac{1}{2} ln (\frac{x + 4 + x ^{2}}{2})) + C

x \to 9 lim (7 x - 4)

f (x) = (x - 3) (4 x + 2)

f (x) = 2 sin^{2} (3 x) + 2 cos^{2} (3 x)

\int_{- 2}^{2} x + 2 d x

f (x) = (\frac{x - 4}{x + 4})^{4}

What is the integral of 5sqrt(x^2+4) ?
The integral of 5sqrt(x^2+4) is 20(1/8 xsqrt(4+x^2)+1/2 ln((|x+sqrt(4+x^2)|)/2))+C