What is the integral of sqrt(5x^2+1) ?

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

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integral of sqrt(5x^2+1)

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{5} sec^{3} (u) d u

= \int \frac{1}{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{1}{5} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

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

Substitute back

u = arctan (5 x)

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{a e ^{x} + b}{a e ^{x} - b} d x

(9 x y^{2} - 4 y) + (9 x^{2} y - 4 x) y^{^{'}} = 0

\int (csc (3 x) - cot (3 x))^{2} d x

y^{^{''}} - 2 y^{^{'}} = sin (x), y (0) = 1, y^{^{'}} (0) = 2

\int t^{2} + 4 d t

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