integral of 8/(4*t^2+1)^2 — step-by-step solution

Let's compute the integral:

\int \frac{8}{{(4 t^{2} + 1)}^{2}} d t

The denominator

4 t^{2} + 1

is strictly positive on ℝ. Let

u = 2 t

, so

du = 2 dt

The integral becomes

4

times

\int \frac{1}{{(1 + u^{2})}^{2}} d u

The quotient rule and the derivative of atan give

{(\frac{u}{1 + u^{2}} + \arctan (u))}^{'}

2 \cdot \frac{1}{{(1 + u^{2})}^{2}}

: an antiderivative of

\frac{1}{{(1 + u^{2})}^{2}}

\frac{\frac{u}{1 + u^{2}} + \arctan (u)}{2}

Multiply this antiderivative by

4

, then substitute

u

back:

2 \cdot (\frac{2 t}{1 + {(2 t)}^{2}} + \arctan (2 t))

ⓘ The power of a product is the product of the powers:

{(2 t)}^{2}

2^{2} \cdot t^{2}

■Expand by multiplying each term inside the parentheses:

2 \cdot (\frac{2 t}{1 + 2^{2} \cdot t^{2}} + \arctan (2 t))

2 \cdot \frac{2 t}{1 + 2^{2} \cdot t^{2}} + 2 \cdot \arctan (2 t)

2 \cdot \frac{2 t}{1 + 2^{2} \cdot t^{2}} + 2 \cdot \arctan (2 t)

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

2 \cdot \frac{2 t}{1 + 4 t^{2}} + 2 \cdot \arctan (2 t)

ⓘ Multiplying a fraction by a number multiplies its numerator:

2 \cdot \frac{2 t}{1 + 4 t^{2}}

\frac{2 \cdot 2 \cdot t}{1 + 4 t^{2}}

\frac{4 t}{1 + 4 t^{2}}

\frac{4 t}{1 + 4 t^{2}} + 2 \cdot \arctan (2 t)

▶Indefinite integrals are defined up to an additive constant C, so this yields:

\frac{4 t}{4 t^{2} + 1} + 2 \cdot \arctan (2 t) + C