Integration Questions

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

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