Integral of 7/(1+x^2) dx?

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Integral of 1/(1+x^2) would be inverse tangent(x).
so the integral of 7/(1+x^2) would be 7*(inverse tangent(x)).
from 1 to 2:
7*(inverse tan(2) - inverse tan(1)) = 7*(63-45) = 7*18 = 126 Degree = 0.7pi...Show more

∫ 1/(1 + x^2) dx = arctan(x) + C is a standard integral (you can derive this via the trigonometric substitution u = tan(x) ==> du = sec^2(x) dx if needed), so we have:
∫ 7/(1 + x^2) dx (from x=1 to 2) = 7 ∫ 1/(1 + x^2) dx (from x=1 to 2)
= 7arctan(x) (evaluated from x=1 to 2), by the FTC part 2 and the above integral
= 7[arctan(2) - arctan(1)]
= 7arctan(2) - 7π/4, by expanding (note: tan(π/4) = 1 ==> arctan(1) = π/4).

I hope this helps!...Show more

Trig substitution.

x = tanø
dx = sec^2ø dø

7 ∫ (sec^2ø)dø / (1 + tan^2ø)

Note that 1 + tan^2ø = sec^2ø by the pythagorean identity:

7 ∫ sec^2ø dø / sec^2ø

7 ∫ 1 dø
7ø

But we want it in terms of x.
So look back at the original substitution:
x = tanø

That means that ø = tan^-1(x)

Therefore:
7ø
is the same as:
7tan^-1(x)

Now evaluate from 1 to 2:
7[tan^-1(2) - tan^-1(1)]...Show more