To determine

To evaluate:

The integral ∫04116+t2 dt.

Answer

∫04116+t2 dt= π16.

Explanation

Given:

∫04116+t2 dt

Formulae used: The integral of the expression ∫bc1a2+x2 dt=1a[tan−1(ca)−tan−1(ba)]

Consider f(x)=116+t2

Hence, use the formulae of the integral ∫bc1a2+x2 dt=1a[tan−1(ca)−tan−1(ba)]

f(x)=116+t2=1(4)2+t2

Now according to the given expression, the integral of the function from t=0 to t=4.

∫04116+t2 dt=∫041(4)2+t2 dt=14tan−1(t4)|04=14[tan−1(1)−tan−1(0)]=π16

Conclusion: Hence the integral of the expression ∫04116+t2 dt is π16.