What is the integral of 64qsqrt(q^2+4) ?

The integral of 64qsqrt(q^2+4) is 64/3 (q^2+4)^{3/2}+C

Popular >

integral of 64qsqrt(q^2+4)

\int 64 q q^{2} + 4 d q

\frac{64}{3} (q^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int 64 q q^{2} + 4 d q

Take the constant out:

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

= 64 \cdot \int q q^{2} + 4 d q

Apply u-substitution

: \int u^{2} d u

= 64 \cdot \int u^{2} d u

Apply the Power Rule

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

= 64 \cdot \frac{u ^{3}}{3}

Substitute back

u = q^{2} + 4

= 64 \cdot \frac{( q ^{2} + 4 ) ^{3}}{3}

Simplify

64 \cdot \frac{( q ^{2} + 4 ) ^{3}}{3} : \frac{64}{3} (q^{2} + 4)^{\frac{3}{2}}

= \frac{64}{3} (q^{2} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{64}{3} (q^{2} + 4)^{\frac{3}{2}} + C

\int \frac{1}{4 x ^{2} + 4 x - 3} d x

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

\int \frac{1}{2 y - 3} d y

\int \frac{3 x - 2}{( x + 1 ) ^{2} ( x - 1 )} d x

\int sin^{2} (w x) d x

What is the integral of 64qsqrt(q^2+4) ?
The integral of 64qsqrt(q^2+4) is 64/3 (q^2+4)^{3/2}+C