What is the integral of 3tsqrt(t^2+8) ?

The integral of 3tsqrt(t^2+8) is (t^2+8)^{3/2}+C

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integral of 3tsqrt(t^2+8)

\int 3 t t^{2} + 8 d t

(t^{2} + 8)^{\frac{3}{2}} + C

Solution steps

\int 3 t t^{2} + 8 d t

Take the constant out:

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

= 3 \cdot \int t t^{2} + 8 d t

Apply u-substitution

: \int u^{2} d u

= 3 \cdot \int u^{2} d u

Apply the Power Rule

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

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

Substitute back

u = t^{2} + 8

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

Simplify

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

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

Add a constant to the solution

= (t^{2} + 8)^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (cos^{6} (y^{7} x^{8}))

x \to 0 lim (x^{3} sin (x^{1}))

arcsec (6 x)

\int \frac{1}{x ( x ^{2} - 100 ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (ln (x^{0.6} y^{0.4}))

What is the integral of 3tsqrt(t^2+8) ?
The integral of 3tsqrt(t^2+8) is (t^2+8)^{3/2}+C