What is the integral of sqrt(6+7t^2) ?

The integral of sqrt(6+7t^2) is 1/2 tsqrt(7t^2+6)+3/(sqrt(7))ln|sqrt(7/6)t+sqrt((7t^2+6)/6)|+C

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integral of sqrt(6+7t^2)

\int 6 + 7 t^{2} d t

\frac{1}{2} t 7 t^{2} + 6 + \frac{3}{7} ln \frac{7}{6} t + \frac{7 t ^{2} + 6}{6} + C

Solution steps

\int 6 + 7 t^{2} d t

Apply Trigonometric Substitution

: \int \frac{6}{7} sec^{3} (u) d u

= \int \frac{6}{7} sec^{3} (u) d u

Take the constant out:

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

= \frac{6}{7} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{6}{7} (\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{6}{7} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{7}{6} t)

= \frac{6}{7} \frac{sec ^{2} ( arctan ( \frac{7}{6} t ) ) sin ( arctan ( \frac{7}{6} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{7}{6} t)) + sec (arctan (\frac{7}{6} t))

Simplify

\frac{6}{7} \frac{sec ^{2} ( arctan ( \frac{7}{6} t ) ) sin ( arctan ( \frac{7}{6} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{7}{6} t)) + sec (arctan (\frac{7}{6} t)) : \frac{1}{2} t 7 t^{2} + 6 + \frac{3}{7} ln \frac{7}{6} t + \frac{7 t ^{2} + 6}{6}

= \frac{1}{2} t 7 t^{2} + 6 + \frac{3}{7} ln \frac{7}{6} t + \frac{7 t ^{2} + 6}{6}

Add a constant to the solution

= \frac{1}{2} t 7 t^{2} + 6 + \frac{3}{7} ln \frac{7}{6} t + \frac{7 t ^{2} + 6}{6} + C

x \to 1 + lim (\frac{( x )}{ln ( x )})

\frac{d y}{d x} = 7 x e^{4 y}

\int x (x - 1)^{8} d x

4 x (3 - 2 x)^{5}, at 2

\int (a + b x)^{n} d x

What is the integral of sqrt(6+7t^2) ?
The integral of sqrt(6+7t^2) is 1/2 tsqrt(7t^2+6)+3/(sqrt(7))ln|sqrt(7/6)t+sqrt((7t^2+6)/6)|+C