You could easily pull off a trig-sub in this problem. You know the familiar Pythagorean Theorem a² + b² = c²? Well you can basically pretend the square root in this integral is part of that. Notice how √(a² + b²) = c , therefore that whole root expression represents the hypotenuse of our right triangle. The square root of the inner values are our legs, conveniently enough we have been given perfect squares.
At this point I recommend you actually draw the triangle and label the hypotenuse and the legs of value x and 2 - where you put these legs does not matter. Now recall basic trigonometry, such as Tanθ, Secθ, etc. Try to decipher which would make the least complicated substitution. For instance, you wouldn't want to pick something crazy like Cscθ = √( x²+4)/x . In this case, you'll want to pick something such as:
Tanθ = x/2
2Tanθ = x
You have to take the derivative of each side so we know what to replace dx with:
Sec²θ dθ = 1/2 dx
2Sec²θ dθ = dx
***************************
∫ 1/[ (2Tanθ)²√( (2Tanθ)² + 4) ] * 2Sec²θ dθ
2 ∫ 1/[ 4Tan²θ√( 4Tan²θ + 4) ] * Sec²θ dθ
Manipulate Tan into its trigonometric identity and simplify:
2 ∫ 1/[ 4Tan²θ√( 4(Sec²θ - 1) + 4) ] * Sec²θ dθ
2 ∫ 1/[ 4Tan²θ√( 4Sec²θ - 4 + 4) ] * Sec²θ dθ
2 ∫ 1/[ 4Tan²θ√( 4Sec²θ) ] * Sec²θ dθ
2 ∫ 1/[ 8Tan²θSecθ ] * Sec²θ dθ
1/4 ∫ Secθ/Tan²θ dθ
Use identities again, to simplify this expression further:
1/4 ∫ (1/Cosθ)(Cosθ/Sinθ)² dθ
1/4 ∫ Cosθ/Sin²θ dθ
****************
u = Sinθ
du = Cosθ dθ
Note: Cosθ dθ takes place in the integral, so replace it with du.
****************
1/4 ∫ 1/u² du
1/4 ∫ u⁻² du
1/4 * u⁻¹/-1
- (1/4)(1/u)
Sub back in your original θ value and simplify 1/Sinθ:
- (1/4)Cscθ
Replace Cscθ with its x-counterpart, based off that triangle you drew earlier. Also put the plus C! :
- (1/4) * √( x²+4)/x + C
If you want, you can combine this into one fraction, but I prefer to keep it separate:
- √( x²+4)/4x + C
There you have it! Since this was a longer problem, I took the derivative of this answer I found and come up with the original integral value, so it assures it is a correct answer. See it here:
Answers & Comments
∫ 1/[ x²√( x² + 4) ] dx
You could easily pull off a trig-sub in this problem. You know the familiar Pythagorean Theorem a² + b² = c²? Well you can basically pretend the square root in this integral is part of that. Notice how √(a² + b²) = c , therefore that whole root expression represents the hypotenuse of our right triangle. The square root of the inner values are our legs, conveniently enough we have been given perfect squares.
At this point I recommend you actually draw the triangle and label the hypotenuse and the legs of value x and 2 - where you put these legs does not matter. Now recall basic trigonometry, such as Tanθ, Secθ, etc. Try to decipher which would make the least complicated substitution. For instance, you wouldn't want to pick something crazy like Cscθ = √( x²+4)/x . In this case, you'll want to pick something such as:
Tanθ = x/2
2Tanθ = x
You have to take the derivative of each side so we know what to replace dx with:
Sec²θ dθ = 1/2 dx
2Sec²θ dθ = dx
***************************
∫ 1/[ (2Tanθ)²√( (2Tanθ)² + 4) ] * 2Sec²θ dθ
2 ∫ 1/[ 4Tan²θ√( 4Tan²θ + 4) ] * Sec²θ dθ
Manipulate Tan into its trigonometric identity and simplify:
2 ∫ 1/[ 4Tan²θ√( 4(Sec²θ - 1) + 4) ] * Sec²θ dθ
2 ∫ 1/[ 4Tan²θ√( 4Sec²θ - 4 + 4) ] * Sec²θ dθ
2 ∫ 1/[ 4Tan²θ√( 4Sec²θ) ] * Sec²θ dθ
2 ∫ 1/[ 8Tan²θSecθ ] * Sec²θ dθ
1/4 ∫ Secθ/Tan²θ dθ
Use identities again, to simplify this expression further:
1/4 ∫ (1/Cosθ)(Cosθ/Sinθ)² dθ
1/4 ∫ Cosθ/Sin²θ dθ
****************
u = Sinθ
du = Cosθ dθ
Note: Cosθ dθ takes place in the integral, so replace it with du.
****************
1/4 ∫ 1/u² du
1/4 ∫ u⁻² du
1/4 * u⁻¹/-1
- (1/4)(1/u)
Sub back in your original θ value and simplify 1/Sinθ:
- (1/4)Cscθ
Replace Cscθ with its x-counterpart, based off that triangle you drew earlier. Also put the plus C! :
- (1/4) * √( x²+4)/x + C
If you want, you can combine this into one fraction, but I prefer to keep it separate:
- √( x²+4)/4x + C
There you have it! Since this was a longer problem, I took the derivative of this answer I found and come up with the original integral value, so it assures it is a correct answer. See it here:
https://www.symbolab.com/solver/step-by-step/%5Cfr...
Set x = 2 tan(u),
Then
√(x² + 4) = 2 sec(u)
dx = 2 sec²(u) du
and therefore the integral becomes
∫ [1 / (4 tan²(u) * 2 sec(u)) * 2 sec²(u) du
. = ∫ 1/4 cot(u) csc(u) du
. = -1/4 csc(u) + C
. = -1/4 √(x²+4) / x + C
. = -√(x²+4) / (4x) + C