What is the integral of y^2sqrt(4y^2+1) ?

The integral of y^2sqrt(4y^2+1) is 1/64 (-2ysqrt(1+4y^2)-ln|2y+sqrt(1+4y^2)|+4y(4y^2+1)^{3/2})+C

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integral of y^2sqrt(4y^2+1)

\int y^{2} 4 y^{2} + 1 d y

\frac{1}{64} (- 2 y 1 + 4 y^{2} - ln 2 y + 1 + 4 y^{2} + 4 y (4 y^{2} + 1)^{\frac{3}{2}}) + C

Solution steps

\int y^{2} 4 y^{2} + 1 d y

Apply Trigonometric Substitution

: \int \frac{1}{8} sec^{3} (u) tan^{2} (u) d u

= \int \frac{1}{8} sec^{3} (u) tan^{2} (u) d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= \frac{1}{8} \cdot \int - sec^{3} (u) + sec^{5} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= \frac{1}{8} (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Substitute back

u = arctan (2 y)

= \frac{1}{8} (- (\frac{1}{2} sec (arctan (2 y)) tan (arctan (2 y)) + \frac{1}{2} ln ∣ tan (arctan (2 y)) + sec (arctan (2 y)) ∣) + \frac{sec ^{4} ( arctan ( 2 y ) ) sin ( arctan ( 2 y ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (2 y)) tan (arctan (2 y)) + \frac{1}{2} ln ∣ tan (arctan (2 y)) + sec (arctan (2 y)) ∣))

Simplify

\frac{1}{8} (- (\frac{1}{2} sec (arctan (2 y)) tan (arctan (2 y)) + \frac{1}{2} ln ∣ tan (arctan (2 y)) + sec (arctan (2 y)) ∣) + \frac{sec ^{4} ( arctan ( 2 y ) ) sin ( arctan ( 2 y ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (2 y)) tan (arctan (2 y)) + \frac{1}{2} ln ∣ tan (arctan (2 y)) + sec (arctan (2 y)) ∣)) : \frac{1}{64} (- 2 y 1 + 4 y^{2} - ln 2 y + 1 + 4 y^{2} + 4 y (4 y^{2} + 1)^{\frac{3}{2}})

= \frac{1}{64} (- 2 y 1 + 4 y^{2} - ln 2 y + 1 + 4 y^{2} + 4 y (4 y^{2} + 1)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{64} (- 2 y 1 + 4 y^{2} - ln 2 y + 1 + 4 y^{2} + 4 y (4 y^{2} + 1)^{\frac{3}{2}}) + C

\int_{0}^{1} x - x d x

\frac{d x}{d t} = sin (2 π (t + 3)), x (3) = 1

\frac{d}{d x} (cos (2 π (1 - x)))

y = 2 x^{2}, (6, 10)

y^{^{''}} + 2 y^{^{'}} - 15 y = 0

What is the integral of y^2sqrt(4y^2+1) ?
The integral of y^2sqrt(4y^2+1) is 1/64 (-2ysqrt(1+4y^2)-ln|2y+sqrt(1+4y^2)|+4y(4y^2+1)^{3/2})+C