What is the integral of 2tsqrt(1+4t^2) ?

The integral of 2tsqrt(1+4t^2) is 1/6 (1+4t^2)^{3/2}+C

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integral of 2tsqrt(1+4t^2)

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

\frac{1}{6} (1 + 4 t^{2})^{\frac{3}{2}} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int t 1 + 4 t^{2} d t

Apply u-substitution

: \int \frac{u}{8} d u

= 2 \cdot \int \frac{u}{8} d u

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 2 \cdot \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 4 t^{2}

= 2 \cdot \frac{1}{8} \cdot \frac{2}{3} (1 + 4 t^{2})^{\frac{3}{2}}

Simplify

2 \cdot \frac{1}{8} \cdot \frac{2}{3} (1 + 4 t^{2})^{\frac{3}{2}} : \frac{1}{6} (1 + 4 t^{2})^{\frac{3}{2}}

= \frac{1}{6} (1 + 4 t^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (1 + 4 t^{2})^{\frac{3}{2}} + C

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What is the integral of 2tsqrt(1+4t^2) ?
The integral of 2tsqrt(1+4t^2) is 1/6 (1+4t^2)^{3/2}+C