Solution : The integration of \(e^x\) with respect to x is \(e^x\) + C. Since \(d\over dx\) \(e^x\) = \(e^x\) dx On integrating both sides, we get \(\int\) \(e^x\) dx = \(e^x\) Hence, the integration of \(e^x\) is \(e^x\) + C
Solution : \(\int\) \(x^2 + x – 1\over x^2 – 1\) dx = \(\int\) (\(x^2 – 1\over x^2 – 1\) + \(x\over x^2 – 1\))dx = \(\int\) 1 dx + \(\int\) \(x\over x^2 – 1\)) dx Let \(x^2 – 1\) = t \(\implies\) 2x dx = dt = x + \(\int\) \(dt\over 2t\) = x …
Integrate \(x^2 + x – 1\over x^2 – 1\) with respect to x. Read More »
Solution : We have, I = \(\int\) log cos x dx By using integraton by parts, I = \(\int\) 1.log cos x dx Taking log cos x as first function and 1 as second function. Then, I = log cos x \(\int\) 1 dx – \(\int\) { \({d\over dx}\) (log cos x) \(\int\) 1 dx …
Solution : We have, I = \(\int\) \(log {1\over x}\) dx I = \(\int\) \(log 1 – log x\) dx = \(\int\) (-log x) dx By using integration by parts formula, Let I = -(\(\int\) log x .1) dx where log x is the first function and 1 is the second function according to ilate …
Solution : We have, I = \(\int\) \(1\over x log x\) dx Put log x = t \(\implies\) \(1\over x\) dx = dt I = \(\int\) \(1\over t\) dt I = log | t | + C I = log |log x| + C Hence, the integration of \(1\over x log x\) is log (log …
Solution : We have, I = \(\int\) x log x dx By using integration by parts, And taking log x as first function and x as second function. Then, I = log x { \(\int\) x dx } – \(\int\) { \({d\over dx}(log x) \times \int x dx\) } dx I = (log x) \(x^2\over …
Solution : We have, I = \((log x)^2\) . 1 dx, Then , where \((log x)^2\) is the first function and 1 is the second function according to ilate rule, I = \((log x)^2\) { \(\int\) 1 dx} – \(\int\) {\(d\over dx\) \((log x)^2\) . \(\int\) 1 dx } dx I = \((log x)^2\) x …
Solution : We have, I = \(sec^{-1}\sqrt{x}\) dx Let x = \(sec^2t\) dx = \(2sec^2 t tan t\) dt I = t.\(2sec^2 t tan t\) dt u = t and v = \(tan^2 t\) I = \(\int\) u.dv = u.v – \(\int\) v.du = \(t.tan^2 t\) – \(\int\) \(tan^2 t\) dt I = \(t.tan^2 t\) …
Solution : We have, I = \(cos^{-1}\sqrt{x}\) . 1 dx By Applying integration by parts, Taking \(cos^{-1}\sqrt{x}\) as first function and 1 as second function. Then I = \(cos^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(cos^{-1}\sqrt{x}\) \(\int\) 1 dx } dx I = x\(cos^{-1}\sqrt{x}\) – \(\int\) \(-1\over 2\sqrt{(1-x)}\sqrt{x}\) . x dx I = x\(cos^{-1}\sqrt{x}\) – …
Solution : We have, I = \(\int\) \(x cos^{-1} x\) dx By using integration by parts formula, I = \(cos^{-1} x\) \(x^2\over 2\) – \(\int\) \(-1\over \sqrt{1 – x^2}\) \(\times\) \(x^2\over 2\) dx I = \(x^2\over 2\) \(cos^{-1} x\) – \(1\over 2\) \(\int\) \(-x^2\over \sqrt{1 – x^2}\) dx = \(x^2\over 2\) \(cos^{-1} x\) – \(1\over …
