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

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

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integral of sqrt(4+5x^2)

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{4}{5} sec^{3} (u) d u

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

Take the constant out:

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

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

Apply Integral Reduction

= \frac{4}{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) ∣

= \frac{4}{5} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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

\int_{0}^{e} x ln (x) d x

y^{^{'}} = (x - 2) (y - 2)

f (x) = - x + 5 x^{- \frac{1}{2}} - 3, (1, 1)

\frac{d y}{d x} = \frac{( x - 3 )}{y - 3}

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