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

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

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integral of sqrt(5+4x^2)

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

\frac{1}{4} (2 x 4 x^{2} + 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} + 5}{5} - \frac{ln ( 5 )}{2})) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{5}{2} sec^{3} (u) d u

= \int \frac{5}{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{5}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{5}{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{5}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{2}{5} x)

= \frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} x ) ) sin ( arctan ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x))

Simplify

\frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} x ) ) sin ( arctan ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x)) : \frac{1}{4} (2 x 4 x^{2} + 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} + 5}{5} - \frac{ln ( 5 )}{2}))

= \frac{1}{4} (2 x 4 x^{2} + 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} + 5}{5} - \frac{ln ( 5 )}{2}))

Add a constant to the solution

= \frac{1}{4} (2 x 4 x^{2} + 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} + 5}{5} - \frac{ln ( 5 )}{2})) + C

ln (2^{x} - 4)

\frac{\partial}{\partial y} (3 cos (x) - sin (x) y)

y = (25 - 4 x^{2})^{2}, x = - \frac{5}{2}, x = \frac{5}{2}

\int \frac{5 x + 3}{x ^{2} + 1} d x

n = 1 \sum \infty 1 9^{n} x^{n} n!

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