Integral ( sqrt(4x^2 + 1) dx ) To solve this, you need to use trig substitution. Let x = (1/2)tan(t) dx = (1/2)sec^2(t) dt. Skipping details, after the substitution you should obtain Integral ( sqrt( tan^2(t) + 1 ) (1/2)sec^2(t) dt ) Which should simplify to (1/2) Integral ( sqrt( sec^2(t) ) sec^2(t) dt ) (1/2) Integral ( sec(t) sec^2(t) dt ) (1/2) Integral ( sec^3(t) dt ) To solve this tricky integral, you have to use parts. Use the information below to get your answer. ∫sec³(x) dx First, split sec³(x) into sec(x) and sec^2(x). ∫ sec(x) sec²(x) dx Use integration by parts. Let u = sec(x). dv = sec²(x) dx du = sec(x)tan(x) dx. v = tan(x) sec(x)tan(x) - ∫sec(x)tan²(x)dx Use the identity tan²(x) = sec²(x) - 1 sec(x)tan(x) - ∫sec(x)[sec²(x) - 1] dx Distribute the sec(x). sec(x)tan(x) - ∫(sec³(x) - sec(x)) dx Separate into two integrals, sec(x)tan(x) - [∫sec³(x)dx - ∫sec(x)dx] One of the integrals worth remembering is ∫sec(x)dx = ln|sec(x) + tan(x)|. Substituting that, we get sec(x)tan(x) - [∫sec³(x)dx - ln|sec(x) + tan(x)|] Distribute the minus sign. sec(x)tan(x) - ∫sec³(x) dx + ln|sec(x) + tan(x)| At this point, it would appear that we've gone in a circle; that isn't the case, because if we look at the whole equation again, ∫sec³(x)dx = sec(x)tan(x) - ∫sec³(x)dx + ln|sec(x) + tan(x)| What we're going to do at this point is *ADD* ∫sec³(x)dx to both sides as if it were a variable. This eliminates the -∫sec³(x)dx on the right hand side, and makes 2 of them on the left hand side. 2∫sec³(x)dx = sec(x)tan(x) + ln|sec(x) + tan(x)| Now, we multiply both sides by (1/2), to get rid of the 2 on the left hand side. ∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| And, don't forget to add a constant. Our final answer is then: ∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C
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∫ ln x dx = x ln x – x + C
You do this using integration by parts.
let u = ln x, dv = dx, and so du = 1/x dx and v = x
∫ ln x dx = uv – ∫ v du = x ln x - ∫ x (1/x) dx =
x ln x – ∫ 1 dx = x ln x – x + C
Integral Of Ln U
as mentioned, you use the integration by parts method.
There's a reason why they call the integral sign "the snake of pain"
can we integrate the ln(x)???...im not sure,i'm just started learning integration
Integration by parts is explained with a link in the source.
Integrate the original integrand by parts:
∫ lnu du
Let f'(u) = 1
f(u) = u
Let g(u) = lnu
g'(u) = 1 / u
∫ f'(u)g(u) du = f(u)g(u) - ∫ f(u)g'(u) du
∫ lnu du = ulnu - ∫ 1 du
∫ lnu du = ulnu - u + C
∫ lnu du = u(lnu - 1) + C
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Integral ( sqrt(4x^2 + 1) dx ) To solve this, you need to use trig substitution. Let x = (1/2)tan(t) dx = (1/2)sec^2(t) dt. Skipping details, after the substitution you should obtain Integral ( sqrt( tan^2(t) + 1 ) (1/2)sec^2(t) dt ) Which should simplify to (1/2) Integral ( sqrt( sec^2(t) ) sec^2(t) dt ) (1/2) Integral ( sec(t) sec^2(t) dt ) (1/2) Integral ( sec^3(t) dt ) To solve this tricky integral, you have to use parts. Use the information below to get your answer. ∫sec³(x) dx First, split sec³(x) into sec(x) and sec^2(x). ∫ sec(x) sec²(x) dx Use integration by parts. Let u = sec(x). dv = sec²(x) dx du = sec(x)tan(x) dx. v = tan(x) sec(x)tan(x) - ∫sec(x)tan²(x)dx Use the identity tan²(x) = sec²(x) - 1 sec(x)tan(x) - ∫sec(x)[sec²(x) - 1] dx Distribute the sec(x). sec(x)tan(x) - ∫(sec³(x) - sec(x)) dx Separate into two integrals, sec(x)tan(x) - [∫sec³(x)dx - ∫sec(x)dx] One of the integrals worth remembering is ∫sec(x)dx = ln|sec(x) + tan(x)|. Substituting that, we get sec(x)tan(x) - [∫sec³(x)dx - ln|sec(x) + tan(x)|] Distribute the minus sign. sec(x)tan(x) - ∫sec³(x) dx + ln|sec(x) + tan(x)| At this point, it would appear that we've gone in a circle; that isn't the case, because if we look at the whole equation again, ∫sec³(x)dx = sec(x)tan(x) - ∫sec³(x)dx + ln|sec(x) + tan(x)| What we're going to do at this point is *ADD* ∫sec³(x)dx to both sides as if it were a variable. This eliminates the -∫sec³(x)dx on the right hand side, and makes 2 of them on the left hand side. 2∫sec³(x)dx = sec(x)tan(x) + ln|sec(x) + tan(x)| Now, we multiply both sides by (1/2), to get rid of the 2 on the left hand side. ∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| And, don't forget to add a constant. Our final answer is then: ∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C