What is the integral of t/(sqrt(at^2+b)) ?

The integral of t/(sqrt(at^2+b)) is (sqrt(b+at^2))/a+C

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integral of t/(sqrt(at^2+b))

\int \frac{t}{a t ^{2} + b} d t

\frac{b + a t ^{2}}{a} + C

Solution steps

\int \frac{t}{a t ^{2} + b} d t

Apply u-substitution

: \int \frac{1}{2 a u} d u

= \int \frac{1}{2 a u} d u

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2 a} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = b + a t^{2}

= \frac{1}{2 a} \cdot 2 (b + a t^{2})^{\frac{1}{2}}

Simplify

\frac{1}{2 a} \cdot 2 (b + a t^{2})^{\frac{1}{2}} : \frac{b + a t ^{2}}{a}

= \frac{b + a t ^{2}}{a}

Add a constant to the solution

= \frac{b + a t ^{2}}{a} + C

\frac{d ^{2}}{d x ^{2}} (3 e^{x} cos (x))

(- 3, 2), (6, 3)

\int 30 arctan (x) d x

2 sin (3 t)

x \to 1 lim (a x + b)

What is the integral of t/(sqrt(at^2+b)) ?
The integral of t/(sqrt(at^2+b)) is (sqrt(b+at^2))/a+C