Using the six integration formulas shown above, we get Now, let’s substitute (x + 3/2) = t, so that dx = dt. Let’s substitute (2x 2 + 6x + 5) = t, so that (4x + 6) dx = dt. Next, let’s equate the coefficients of ‘x’ and the constant terms. Let’s understand with the help of some examples: Solved Problems for You Example 1: Find ∫ dx ‘A’ and ‘B’ are thus obtained and hence the integral is reduced to one of the known forms. To determine ‘A’ and ‘B’, we equate from both sides the coefficients of x and the constant terms. (px + q) = A d/dx (ax 2 + bx + c) + B = A (2ax + b) + B To solve this, we must find constants A and B such that, Integral ∫ dx, where p, q, a, b, and c are constants. Remember, you can also solve ∫ dx / √ (ax 2 + bx + c) in a similar manner. This can be evaluated using one / more of the six integration formulas shown above. Therefore,Īx 2 + bx + c = a (t 2 ± k 2) … where the + or – depends on the sign of (c/a – b 2/4a 2). Also, substitute (c/a – b 2/4a 2) = ±k 2. Now, let’s substitute (x + b/2a) = t, so that dx = dt. Now, let’s apply these standard integration formulas to obtain formulae that are applied directly to evaluate other integrals. = log |x + √(x 2 + a 2)| + C … where C = C 1 – log |a| ∫ dx / √ (x 2 + a 2) = ∫ a sec 2 t dt / √ (a 2 tan 2 t + a 2) Let’s substitute x = a tan t, so that dx = a sec 2 t dt. ∫ dx / √ (a 2 – x 2) = ∫ a cos t dt / √ (a 2 – a 2 sin 2 t) Let’s substitute x = a sin t, so that dx = a cos t dt. ∫ dx / √ (x 2 – a 2) = ∫ sec t dt = log |sec t + tan t| + C 1 ∫ dx / √ (x 2 – a 2) = ∫ a sec t tan t dt / √ (a 2 sec 2 t – a 2)

Let’s substitute x = a sec t, so that dx = a sec t tan t dt. Let’s substitute x = a tan t, so we have dx = a sec 2 t dt. = 1/2a log |(x – a) / (x + a)| + C Browse more Topics Under Integrals Proof of the Above Six Standard Integration Formulas 1. Look at the following integration formulas

Integral formula sheet